Core Connections Integrated 3
Core Connections Integrated 3
1st Edition
Brian Hoey, Judy Kysh, Leslie Dietiker, Tom Sallee
ISBN: 9781603283939
Textbook solutions

All Solutions

Page 225: Questions

Exercise 1
Step 1
1 of 2
Assuming the unknown number is $x$, then,

$10 cdot (x+4)=-70$          (Adding 4 to the number and multiplying the sum by 10).

$x+4=-7$          (Divide each side by 10)

$x=-7-4$          (Subtract 4 from each side)

$$
x=-11
$$

Result
2 of 2
$$
x=-11
$$
Exercise 2
Step 1
1 of 5
a-\
The following table shows the input and output of the machine.\
begin {center}
begin{tabular}{|r|r|}
hline
Input & Output \
hline
3 & 7 \
hline
4 & 9 \
hline
-3 & -5 \
hline
end{tabular}
end {center}
The machine multiplies the input by 2 and add 1 to the sum to generate the result.\
Step 2
2 of 5
b-

If the machine is reversing its function, the expected outputs for the different inputs will be.

For pulling back 7 the value that comes out is 3.

For pulling back 9 the value that comes out is 4.

For pulling back -5 the value that comes out is -3.

Step 3
3 of 5
c-\
The table below demonstrates the inputs and outputs of the backwards function.\
begin {center}
begin{tabular}{|r|r|}
hline
x & y \
hline
7 & 3 \
hline
9 & 4 \
hline
-5 & -3 \
hline
end{tabular}
end {center}
Anita’s backward function is subtracting from the input and dividing the difference by 2.\
Step 4
4 of 5
d-

Anita’s original function machine is:

$y=2x+1$

Anita’s backward function machine is:

$y=dfrac {x-1}{2}$

Result
5 of 5
a-          The machine multiplies the input by 2 and add 1 to the sum to generate the result.

b-          The expected outputs for the different inputs of the reverse function will be.

3 for input 7, 4 for input 9 and -3 for input -5
c-          Anita’s backward function is subtracting from the input and dividing the difference by 2.

d-          Anita’s original function machine is:          $y=2x+1$

Anita’s backward function machine is:          $y=dfrac {x-1}{2}$

Exercise 3
Step 1
1 of 5
a-

$h(4)=5 cdot 4+2=22$

22 is the number that should be pulled up into the machine in order to have a 4 come out.

Step 2
2 of 5
b-

Keiko’s machine must subtract 2 from 17 and divide the difference by 5 to undo the original function.

Step 3
3 of 5
c-

The inverse function is:

$h^{-1}(x)=dfrac {x-2}{5}$

Step 4
4 of 5
d-

Let’s check the original function $h(x)$ and the inverse function $h^{-1}(x)$ for $x=4$

$h(4)=5 cdot 4+2=22$

$h^{-1}(22)=dfrac {22-2}{5}=dfrac {20}{5}=4$

Result
5 of 5
a-          22 is the number that should be pulled back into the machine in order to have a 4 come out.

b-          Keiko’s machine must subtract 2 from 17 and divide the difference by 5 to undo the original function.

c-          The inverse function is:          $h^{-1}(x)=dfrac {x-2}{5}$

d-          $h(4)=5 cdot 4+2=22$

$h^{-1}(22)=dfrac {22-2}{5}=dfrac {20}{5}=4$

Exercise 4
Step 1
1 of 2
Exercise scan
Step 2
2 of 2
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Exercise 5
Step 1
1 of 3
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Step 2
2 of 3
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Step 3
3 of 3
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Exercise 6
Step 1
1 of 3
$g(x)=x^3-5$

$$
g^{-1}(x)=sqrt[3]{x+5}
$$

We are given the function and its inverse:
Step 2
2 of 3
begin{center}
begin{tabular}{|| c|c|c| c||}
hline
$x$ & $y=g(x)$ & $(x,y)$ & $(y,x)$ \ [0.5ex]
hline
-2 &-13 & (-2,-13) & (-13,-2) \
hline
-1 &-6 & (-1,-6) & (-1,-6) \
hline
0 &-5 & (0,-5) & (-5,0) \
hline
1 &-4 & (1,-4) & (-4,1) \
hline
2 &3 & (2,3) & (3,2) \
hline
3 &22 & (3,22) & (22,3) \[1ex]
hline
end{tabular}
end{center}
We make a table of values:
Step 3
3 of 3
Exercise scan
We graph both functions:
Exercise 7
Step 1
1 of 1
Exercise scan
Exercise 8
Step 1
1 of 3
Work as shown below, follow the steps:

$color{#c34632} text{$y=dfrac{1}{2}x-3$}$

Graph this line by plotting points:

Find at least two ordered pairs $color{#c34632} text{$,,(x,y),,$}$ that satisfy this equation:

$-$For $color{#4257b2} text{$,x=0,$}$ we have: $,,,y=dfrac{1}{2}(0)-3Rightarrow color{#4257b2} text{$y=-3$}$

$-$For $color{#4257b2} text{$,y=0,$}$ we have: $,,, 0=dfrac{1}{2}x-3Rightarrow color{#4257b2} text{$x=6$}$

Plot the points $color{#4257b2} text{$,,(0 , -3),,$}$ and $color{#4257b2} text{$,,(6,0),,$}$ and connect them to get the graph of this equation as shown in the picture below:

Exercise scan

Step 2
2 of 3
In order to graph the inverse relation plot the points $color{#4257b2}text{$,,(-3 ,0),,$}$ and $color{#4257b2} text{$,,(0,6),,$}$ and connect them as shown in the picture:

Exercise scan

Step 3
3 of 3
a) Find the inverse of the given equation as shown, follow the steps:

$y=dfrac{1}{2}x-3qquadqquadqquadqquadqquad$ $color{#c34632} text{[interchange $x$ and $y$]}$

$Rightarrow x=dfrac{1}{2}y-3qquadqquadqquadqquadqquad$ $color{#c34632} text{[solve for $y$]}$

$Rightarrow dfrac{y}{2}=x+3$

$Rightarrow color{#4257b2} text{$y=2x+6$}$

The equation of the inverse function is: $color{#4257b2} text{$quad y=2x+6$}$

b) As we know the graphs of two inverse functions are always symmetric with respect to the line $color{#4257b2} text{$,,y=x,$}$:

This can be illustrated by graphing the line $color{#4257b2} text{$,,y=x,,$}$ along with the two lines above:

Exercise scan

Exercise 9
Step 1
1 of 2
textbf{(a)} Evaluating $A(2)$:
begin{align*}
A(2)&=3^2\
&=9
intertext{textbf{(b)} Let $a$ be the input for to get 81 as output, then:}
3^{a}&=81\
3^{a}&=3^{4}
intertext{Comparing powers both sides:}
a&=4
intertext{textbf{(c)} Let $b$ be the input for to get 8 as output, then:}
3^{b}&=8\
intertext{Taking $log$ both sides:}
log (3^{b})&=log 8\
blog 3&=log 8 & textrm{As, $log x^a=alog x$}\
b&=dfrac{log 8}{log 3}
intertext{Using calculator:}
b&=1.89
end{align*}
Result
2 of 2
(a) 9 (b) 4 (c) 1.89
Exercise 10
Step 1
1 of 6
In this exercise, we need to sketch a couch, determine what a vertical cross-section down the middle of the couch would look like, and then sketch the cross-section.

*How do we determine the shape of a cross-section?*

Step 2
2 of 6
A **cross-section** is the shape created by the intersection of a 3D solid and a plane. A cross-section is therefore a 2D shape that represents the flat surface of a specific slice of the solid.

When thinking about cross-sections, it helps to think about slicing a loaf of bread. The loaf of bread is the 3D solid and the slice of bread is a cross-section.

Step 3
3 of 6
Let’s start by making a sketch of a couch:

$$text{Figure 1: Sketch of a couch}$$

Note: Couches can vary in size and shape so your sketch may have some differences compared to Figure $1$.

Step 4
4 of 6
Next, let’s think about what happens if we slice the couch vertically right down the middle of the couch like so:

Step 5
5 of 6
The shape created would be an L-shape since the seat and backrest of the coach create an L-shape. The armrests and legs of the coach wouldn’t be visible in the cross-section since they are at the ends of the couch. The cross-section would then look like this:

Step 6
6 of 6
**Summary:**

First, we recalled what a cross-section is. Then we sketched a couch. After determining what shape is created by the middle section of the couch, we then sketched the cross-section.

Exercise 11
Step 1
1 of 11
a) We can use the $ln$ function

$$
begin{align*}
1^x &= 5\
ln 1^x &= ln 5
end{align*}
$$

Now we can use the formula

$$
begin{align*}
color{#c34632}{ln a^b = b ln a} tag {*}
end{align*}
$$

Step 2
2 of 11
Then

$$
begin{align*}
x ln 1 &= ln 5\
x &= frac {ln 5}{ln 1}\
x &= frac {1.61}{0} &&text{( $ln 1=0$)}\
x &= infty
end{align*}
$$

The solution of this equation does not exist.

Step 3
3 of 11
b) We will write the given equation in the following form

$$
begin{align*}
sqrt {27^x} &= 81\
27^x &= 81^2\
27^x &= 6561
end{align*}
$$

Now we can use the $ln$ function

$$
ln 27^x = ln 6561
$$

Step 4
4 of 11
We use the formula $(*)$

$$
begin{align*}
x ln 27 &= ln 6561\
x &= frac {ln 6561}{ln 27}\
x &= frac {8.78}{3.29}\
x &= 2.66
end{align*}
$$

The solution of this equation is

$$
x = 2.66
$$

Step 5
5 of 11
c) We can use the $ln$ function

$$
begin{align*}
2^x &= 9\
ln 2^x &= ln 9
end{align*}
$$

We use the formula $(*)$

$$
begin{align*}
x ln 2 &= ln 9\
x &= frac {ln 9}{ln 2}\
x &= frac {2.19}{0.69}\
x &= 3.17
end{align*}
$$

The solution of this equation is

$$
x = 3.17
$$

Step 6
6 of 11
d) This equation we can write

$$
begin{align*}
25^{(x+1)} &= 125^x\
(5^2)^{(x+1)} &= (5^3)^x \
5^{2(x+1)} &= 5^{3x} && text{( $(a^n)^m=a^{n cdot m}$ )}
end{align*}
$$

Step 7
7 of 11
According to the formula

$$
begin{align*}
color{#c34632}{a^b = a^c} tag {**}\
color{#c34632}{b=c}
end{align*}
$$

we obtain

$$
begin{align*}
2( x+1 ) &= 3x\
2x + 2 &= 3x\
2 &= 3x-2x\
x &= 2
end{align*}
$$

The solution of this equation is

$$
x=2
$$

Step 8
8 of 11
e) This equation we can write

$$
begin{align*}
8^x &= 2^5 cdot 4^4\
left( 2^3right)^x &= 2^5 left( 2^2 right)^4\
2^{3x} &= 2^5 cdot 2^8
end{align*}
$$

Step 9
9 of 11
We can use the formula

$$
color{#c34632}{a^b cdot a^c = a^{b+c}}
$$

Now we have

$$
begin{align*}
2^{3x} &= 2^{5+8}\
2^{3x} &= 2^{13}
end{align*}
$$

Step 10
10 of 11
According to $(**)$ we get

$$
begin{align*}
3x &= 13\
x &= frac {13}{3}\
x &= 4.33
end{align*}
$$

The solution of this equation is

$$
x = 4.33
$$

Result
11 of 11
a) $x = infty$

b) $x = 2.66$

c) $x = 3.17$

d) $x = 2$

e) $x = 4.33$

Exercise 12
Step 1
1 of 5
a) Tasha has $$3$. In the first month, Tasha has $2 cdot 3 = $ 6$. In the second month, Tasha has $2 cdot 6 = $ 12$.

Therefore

1. month $= 3 cdot 2$

2. month $= 3 cdot 2 cdot 2 = 3 cdot 2^2$

3. month $= 3 cdot 2 cdot 2 cdot 2 = 3 cdot 2^3$

In the $x$ month, Tasha has

$$
y = 3 cdot 2^x
$$

Step 2
2 of 5
b) Clifton has $$10$.

1. month $= 10 cdot 2$

2. month $= 10 cdot 2 cdot 2 = 10 cdot 2^2$

3. month $= 10 cdot 2 cdot 2 cdot 2 = 10 cdot 2^3$

In the $x$ month, Clifton has

$$
y = 10 cdot 2^x
$$

Step 3
3 of 5
c) The graph of the functions

$$
begin{align*}
y = 3 cdot 2^x\
y = 10 cdot 2^x
end{align*}
$$

Exercise scan

Step 4
4 of 5
Graph function $y=10 cdot 2^x$ grows faster than graph function $y= 3 cdot 2^x$.
Result
5 of 5
a) $y = 3 cdot 2^x$

b) $y = 10 cdot 2^x$

c) Graph function $y=10 cdot 2^x$ grows faster than graph function $y= 3 cdot 2^x$.

Exercise 13
Step 1
1 of 3
The function $y=|x-2|$ is the absolute function translated to the right by 2 units.

The function $y=4-|x|$ is the absolute function reflecte about te $x$-axis and translated up by 4 units.

The enclosed area is colored orange.

Exercise scan

Step 2
2 of 3
Note that the enclosed area is a rectangle with the length of the rectangle being the hypotenuse of a right triangle with rectangular sides of 3 each, and the width of the rectangle being the hypotenuse of a right triangle with rectangular sides of 1 each.

Pythagorean theorem for right triangle with $c$ the hypotenuse:
$$
a^2+b^2=c^2
$$

Using the Pythagorean theorem, we can then determine the length and the width of the rectangle:

$$
text{Length}=c=sqrt{a^2+b^2}=sqrt{3^2+3^2}=sqrt{18}=3sqrt{2}
$$

$$
text{Width }=c=sqrt{a^2+b^2}=sqrt{1^2+1^2}=sqrt{2}
$$

The area of a rectangle is the product of the length and the width of the rectangle.

$$
AREA=text{length}times text{width}=3sqrt{2}times sqrt{2}=3times 2=6
$$

Result
3 of 3
AREA=$6$
Exercise 14
Step 1
1 of 2
If the ticket costs $$11.5$, then the expression

$$
11.5 ( 1.083)^t
$$

can mean that the price of ticket rose by $1.083$ times.

The ticket is growing every month for $8.3 %$.

Result
2 of 2
The ticket is growing every month for $8.3 %$.
Exercise 15
Step 1
1 of 10
$f(x)=0.5x+3$

Exercise scan

a) We are given the function:
Step 2
2 of 10
Exercise scan
We make the graph of the inverse:
Step 3
3 of 10
$g(x)=3(x+2)^2-6$

Exercise scan

We are given the function:
Step 4
4 of 10
Exercise scan
We make the graph of the inverse:
Step 5
5 of 10
$h(x)=dfrac{1}{6}x^3-dfrac{13}{6}x+2$

Exercise scan

We are given the function:
Step 6
6 of 10
Exercise scan
We make the graph of the inverse:
Step 7
7 of 10
b) If we note by $(x,y)$ the coordinates of the function, then the coordinates of the inverse are $(y,x)$ as the inverse is the reflection of the function across the $y=x$ line.
Step 8
8 of 10
begin{center}
begin{tabular}{|| c|c|c| c||}
hline
$x$ & $y=f(x)$ & $(x,y)$ & $(y,x)$ \ [0.5ex]
hline
-2 & 2 & (-2,2) & (2,-2) \
hline
0 & 3 & (0,3) & (3,0) \
hline
2 & 4 & (2,4) & (4,2) \[1ex]
hline
end{tabular}
end{center}
For the $f(x)$ function we built the table of values:
Step 9
9 of 10
begin{center}
begin{tabular}{|| c|c|c| c||}
hline
$x$ & $y=g(x)$ & $(x,y)$ & $(y,x)$ \ [0.5ex]
hline
-3 & -3 & (-3,-3) & (-3,-3) \
hline
-2 & -6 & (-2,-6) & (-6,-2) \
hline
-1 & -3 & (-1,-3) & (-3,-1) \
hline
0 & 6 & (0,6) & (6,0) \[1ex]
hline
end{tabular}
end{center}
For the $g(x)$ function we built the table of values:
Step 10
10 of 10
begin{center}
begin{tabular}{|| c|c|c| c||}
hline
$x$ & $y=h(x)$ & $(x,y)$ & $(y,x)$ \ [0.5ex]
hline
-4 & 0 & (-4,0) & (0,-4) \
hline
-2 & 5 & (-2,5) & (5,-2) \
hline
0 & 2 & (0,2) & (2,0) \
hline
1 & 0 & (1,0) & (0,1) \
hline
2 & -1 & (2,-1) & (-1,2) \
hline
3 & 0 & (3,0) & (0,3) \[1ex]
hline
end{tabular}
end{center}
For the $h(x)$ function we built the table of values:
Exercise 16
Step 1
1 of 8
$$
y=0.5x+3
$$
a) We are given the function:
Step 2
2 of 8
Exercise scan
We draw the line of symmetry for the function $y=0.5x+3$ and its inverse:
Step 3
3 of 8
$$
y=3(x+2)^2-6
$$
We are given the function:
Step 4
4 of 8
Exercise scan
We draw the line of symmetry for the function $y=3(x+2)^2-6$ and its inverse:
Step 5
5 of 8
$$
y=dfrac{1}{6}x^3-dfrac{13}{6}x+2
$$
We are given the function:
Step 6
6 of 8
Exercise scan
We draw the line of symmetry for the function $y=dfrac{1}{6}x^3-dfrac{13}{6}x+2$ and its inverse:
Step 7
7 of 8
$$
y=x
$$
b) The equation of the line of symmetry in each graph is:
Step 8
8 of 8
c) The coordinates of the points $(x,y)$ in the graph of the original function are reflected to $(y,x)$ in the graph of the inverse, which is a reflection across the $y=x$ line.
Exercise 17
Step 1
1 of 6
$$
y=left(dfrac{x}{2}right)^2
$$
a) We graph the function:
Step 2
2 of 6
Exercise scan
We graph the function:
Step 3
3 of 6
Exercise scan
b) We graph the line of symmetry $y=x$:
Step 4
4 of 6
c) We make a carbon copy of the parabola.
Step 5
5 of 6
Exercise scan
d) We got the inverse of the original function:
Step 6
6 of 6
The two graphs show the function and its inverse as they are symmetrical with respect to the line $y=x$.

Because the inverse parabola is not a function, we must work with restrictions: for example the original function restricted to $(0,infty)$ has the inverse the upper half of the inverse curve.

Exercise 18
Step 1
1 of 3
a) We will calculate the domain and rank of the function

$$
begin{align*}
y &= left( frac x2 right)^2\
y &= frac {x^2}{4}
end{align*}
$$

The domain of this function is

$$
D = ( – infty , + infty )
$$

and rank of this function is

$$
R = [ 0, + infty )
$$

Inverse function:

$$
begin{align*}
4y &= x^2\
x &= pm sqrt {4y}\
x &= pm 2 sqrt y
end{align*}
$$

The domain of this function is

$$
D = [0, + infty )
$$

and rank of this function is

$$
R = ( – infty , + infty )
$$

Step 2
2 of 3
b) The domain of the function

$$
y = left( frac x2 right)^2
$$

is rank of the function

$$
x = pm 2 sqrt y
$$

and vice versa.

Result
3 of 3
a) $D = ( – infty , + infty )$

$R = ( – infty, + infty )$

b) The domain of the function

$$
y = left( frac x2 right)^2
$$

is rank of the function

$$
x = pm 2 sqrt y
$$

and vice versa.

Exercise 19
Step 1
1 of 2
$$
y=left(dfrac{x}{2}right)^2
$$
We are given the function:
Step 2
2 of 2
Exercise scan
We use a graphing calculator to draw both the function and its inverse:
Exercise 20
Step 1
1 of 3
(a) Interchange $x$ and $y$ in $y=y(x)$ to obtain the inverse $y^{-1}(x)$;
Step 2
2 of 3
$x=dfrac{y^{2}}{4}$ then $y^{-1}(x)=pmsqrt{4x}=pm2sqrt{x}$. So the equation is $y=pm2sqrt{x}$ This could also be written as $y^{2}=4x$. Check that the inverse function undoes the original function by demonstrating $y(y^{-1}(x))=x$. So here; $y(y^{-1}(x))=left( dfrac{2sqrt{x}}{2}right)^{2}=x$
Result
3 of 3
$$
y=pm2sqrt{x}
$$
Exercise 21
Step 1
1 of 4
(a) The inverse equation is $y=pm2sqrt{x}$ which is not a function because it does not satisfy the vertical line test for a function. Elements in the domain of the inverse are mapped to more than one element in the range.
Step 2
2 of 4
(b) $;$

Exercise scan

Step 3
3 of 4
(b) and (c) Plot shows the part of the inverse; $y=2sqrt{x}$ for $xgeq0$. This satisfies the requirements for a function since we have a bijection; a one to one mapping between elements. The new function does not generate any negative values i.e. $y(x)geq0$ whereas previously $y(x)=pm2sqrt{x}$.

Exercise scan

Result
4 of 4
see graphs and explanations
Exercise 22
Step 1
1 of 4
$$
g(x)=(x-3)^2
$$
We are given the function:
Step 2
2 of 4
$$
[3,infty)
$$
a) We restrict the domain of $g(x)$ so that its inverse is a function: we use the vertex $(3,0$:
Step 3
3 of 4
Exercise scan
b) We graph $g(x)$ on the restricted domain and its inverse:
Step 4
4 of 4
$g(x)=(x-3)^2$

$y=(x-3)^2$

$x=(y-3)^2$

$sqrt x=y-3$

$y=sqrt x+3$

$$
g^{-1}(x)=sqrt x+3, xgeq 0
$$

c) We determine the equation of $g^{-1}(x)$:
Exercise 23
Step 1
1 of 1
Given the graph of a function $f$ we can determine if its inverse is also a function by applying the horizontal line test to the graph of $f$.

According to the horizontal line test, any horizontal line must intersect the graph of $f$ at most in one point.

If there is more than one points of intersection the inverse of $f$ will not be a function.

An example of a function $f$ whose inverse is not a function is the function:

$f(x)=x^2$

The inverse of this function is not a function because the graph of $f$ does not pass the horizontal line test.

Exercise 24
Step 1
1 of 3
Use a graphing calculator for the graph of $f(x)=1+dfrac{2}{x}$(red curves).

Sketch the graph of the inverse function $f^{-1}(x)$(green colour) with symmetry with respect to the line $y=x$.

See the picture below:

Exercise scan

Step 2
2 of 3
a) Find the inverse $f^{-1}$ as shown below:

$y=1+dfrac{2}{x}qquad$ [interchange $x$ and $y$]

$Rightarrow x=1+dfrac{2}{y}qquad$ [silve for $y$]

$Rightarrow x-1=dfrac{2}{y}qquad$ [cross multiplication]

$Rightarrow y=dfrac{2}{x-1}qquad$ [set $y=f^{-1}(x)$]

$$
Rightarrow f^{-1}(x)=dfrac{2}{x-1}
$$

Step 3
3 of 3
b) As we can see from the graph the domain and the range of $f$ is:

$$
D_{f}=(-infty , 0)cup (0,+infty)
$$

$R_{f}=(-infty ,1)cup (1,+infty)$

The domain and the range of $f^{-1}$ is:

$$
D_{f^{-1}}=(-infty ,1)cup (1,+infty)
$$

$$
R_{f^{-1}}=(-infty ,0)cup (0,+infty)
$$

Exercise 25
Step 1
1 of 2
$-$First find the inverse $f^{-1}$ of the given function as shown below:

$f(x)=dfrac{1}{2}(x-1)^3$

$Rightarrow y=dfrac{1}{2}(x-1)^3qquad$ [interchange $x$ and $y$]

$Rightarrow x=dfrac{1}{2}(y-1)^3qquad$ [solve for $y$]

$Rightarrow (y-1)^3=2xqquad$ [take cube roots]

$Rightarrow y-1=sqrt[3]{2x}qquad$ [add $1$]

$Rightarrow y=sqrt[3]{2x}+1qquad$ [set $y=f^{-1}(x)$]

$Rightarrow f^{-1}(x)=sqrt[3]{2x}+1$

$-$The graphs of $f$(red colour) and $f^{-1}$(blue colour) is shown in the pic below.

Exercise scan

Result
2 of 2
$$
f^{-1}=sqrt[3]{2x}+1
$$
Exercise 26
Step 1
1 of 4
a-

$y=3x-8$          (Given)

To find the inverse function:

$y+8=3x$          (Add 8 to each side)

$x=dfrac {y+8}{3}$

The inverse function is:

$$
y=dfrac {x+8}{3}
$$

Step 2
2 of 4
b-

$y=dfrac {1}{2}x+6$          (Given)

To find the inverse function:

$2y=x+12$          (Multiply each side by 2)

$x=2y-12$

The inverse function is:

$$
y=2x-12
$$

Step 3
3 of 4
c-

$y=dfrac {x+6}{2}$          (Given)

$2y=x+6$          (Multiply each side by 2)

$x=2y-6$

The inverse function is:

$$
y=2x-6
$$

Result
4 of 4
a-          $y=dfrac {x+8}{3}$

b-          $y=2x-12$

c-          $y=2x-6$

Exercise 27
Step 1
1 of 3
Exercise scan
If the rotation line is parallel to one of the rectangle’s sides, then the result is a cylinder:
Step 2
2 of 3
Exercise scan
If the rotation line is a diagonal, for example, the result is not a cylinder:
Result
3 of 3
Disagree
Exercise 28
Step 1
1 of 12
In this exercise, we need to determine the values of $a$ and $b$ that make each system of equations true.
Step 2
2 of 12
When solving systems of non-linear equations, we can use the **substitution method**. With this method,
* Solve one of the equations for a variable.
* Substitute the expression you found for the variable into the other equation.
* Solve for the variable.
* Substitute the value of the variable into either equation to find the value of the other variable.

Step 3
3 of 12
Since the equations have exponents, we will also need to use the following exponent properties:
* **Quotient of Powers Property:**
$frac{x^a}{x^b}=x^{a-b}$
* **One-to-One Property:** If $x^a=x^b$, then $a=b$.
Step 4
4 of 12
### a)

Since $b^0=1$ for all $bne0$, we can simplify the first equation to get the value of $a$:
$$begin{align*}
3&=acdot b^0\
3&=acdot1\
3&=a.
end{align*}$$

Step 5
5 of 12
Substitute $a=3$ into the second equation and solve for $b$ using the One-to-One Property:
$$begin{align*}
75&=acdot b^2\
75&=3cdot b^2\
frac{75}{3}&=frac{3cdot b^2}{3}\
25&=b^2\
5^2&=b^2\
5&=b.
end{align*}$$
Step 6
6 of 12
Therefore, the values of $a$ and $b$ that make the system true are $a=3$ and $b=5$.
Step 7
7 of 12
### b)

Solve the first equation for $a$:
$$begin{align*}
18&=acdot b^2\
frac{18}{b^2}&=frac{acdot b^2}{b^2}\
frac{18}{b^2}&=a.
end{align*}$$

Step 8
8 of 12
Substitute $a=frac{18}{b^2}$ into the second equation. Use the Quotient of Powers Property to simplify and solve for $b$:
$$begin{align*}
54&=acdot b^3\
54&=frac{18}{b^2}cdot b^3\
54&=18cdot b^{3-2}\
54&=18b\
frac{54}{18}&=frac{18b}{18}\
3&=b.
end{align*}$$
Step 9
9 of 12
Substitute $b=3$ into either equation and solve for $a$. I have chosen to substitute it into the first equation:
$$begin{align*}
18&=acdot b^2\
18&=acdot 3^2\
18&=acdot9\
frac{18}{9}&=frac{acdot 9}{9}\
2&=a.
end{align*}$$
Step 10
10 of 12
Therefore, the values of $a$ and $b$ that make the system true are $a=2$ and $b=3$.
Step 11
11 of 12
**Summary:**

First, we recalled the steps to solve a system of equations using the substitution method. Then we recalled some applicable exponent properties. Next, we used the substitution method and those properties to solve for the variables.

Result
12 of 12
a) $a=3$ and $b=5$

b) $a=2$ and $b=3$

Exercise 29
Step 1
1 of 2
textbf{(a)} From the information given in question:
$$L(x)=x^2-1$$
$$R(x)=3(x+2)$$
textbf{(b)} Evaluating $R circ L(x)$:
begin{align*}
R circ L(x)&=R(x^2-1)\
&=3(x^2-1+2)\
&=3(x^2+1)\
&=3x^2+3
intertext{Putting $x=3$ in $R circ L(x)$:}
R circ L(3)&=3(3)^2+3\
&=3cdot 9+3\
&= 27+3\
&=30
end{align*}
textbf{(c)} Evaluating $L circ R(x)$:
begin{align*}
L circ R(x)&=L(3(x+2))\
&=(3(x+2))^2-1\
&=9(x+2)^2-1\
&=9(x^2+4x+4)-1\
&=9x^2+36x+36-1\
&=9x^2+36x+35\
intertext{Putting $x=3$ in $L circ R(x)$:}
L circ R(x)&= 9(3)^2+36(3)+35\
&=9cdot 9 +108+35\
&=81+143\
&=224\
end{align*}
It can be observed that result is changed when order of machines is changed.\\
Result
2 of 2
(a) $L(x)=x^2-1$, $R(x)=3(x+2)$ (b) 30 (c) Result is changed
Exercise 30
Solution 1
Solution 2
Step 1
1 of 2
The center of the given circle is $(0,-2)$, now it is shifted 2 units left, then the center will be $(-2,-2)$ and given radius also becomes double,then the equation is
$$(x+2)^2+(y+2)^2=(2r)^2$$
Result
2 of 2
$$(x+2)^2+(y+2)^2=(2r)^2$$
Step 1
1 of 2
x^2+(y+2)^2=(2r)^2
Add 2 to the x side and subtract 5 to the y side. Look at solution below.
Result
2 of 2
(x+2)^2+(y-3)^2=4r^2
Exercise 31
Step 1
1 of 3
The two sloped boundaries are line segments that have eqivalent gradients given by $m=dfrac{5-1}{-3-3}=-dfrac{2}{3}$. The sloped boundary furthest from the origin intercepts the y-axis at $y=3$ and the sloped boundary closest to the origin intercepts the y-axis at $y=-3$. The equations of these two line segments are then $y=-dfrac{2}{3}pm3$.
Step 2
2 of 3
The vertical boundaries are at $x=-3$ and $x=3$
Step 3
3 of 3
Since all boundaries are solid lines the system of inequalities is;
$$
begin{bmatrix}xgeq-3\xleq3\yleq-dfrac{2}{3}x+3\ygeq-dfrac{2}{3}x-3end{bmatrix}
$$
Exercise 32
Step 1
1 of 1
$-$First find the inverse $f^{-1}$ of the given function as shown below:

$f(x)=5(x-2)$

$Rightarrow y=5(x-2)qquad$ [interchange $x$ and $y$]

$Rightarrow x=5(y-2)qquad$ [solve for $y$]

$Rightarrow y-2=dfrac{x}{5}qquad$ [add $2$]

$Rightarrow y=dfrac{x}{5}+2qquad$ [set $y=f^{-1}(x)$]

$Rightarrow f^{-1}(x)=dfrac{x}{5}+2$

$-$The graphs of $f$(red colour) and $f^{-1}$(blue colour) is shown in the pic below.

Exercise scan

Exercise 33
Result
1 of 1
We can’t tell from the graph which is which. because each of them is an inverse for the other.
Exercise 34
Step 1
1 of 7
In this exercise, we are given the graph of a function and need to use it to create the graph of its inverse function. We also need to find the domain and range for both functions and determine the relationships between them.

*When given a point for a function $f(x)$, how can we find a point for $f^{-1}(x)$?*

Step 2
2 of 7
An **inverse function** is a function that “undoes” the operations of the original function $f(x)$. An inverse function is notated as $f^{-1}(x)$. Note that the $-1$ is not an exponent for this notation. Occasionally, the domain of $f(x)$ must be restricted to guarantee that the inverse is a function.

The graphs of a function $f(x)$ and its inverse function $f^{-1}(x)$ are reflections across the line $y=x$. This means for every point $(a,b)$ on the graph of $f(x)$, there must be the point $(b,a)$ on the graph of $f^{-1}(x)$.

Step 3
3 of 7
First, let’s determine some points for the graph of $f(x)$. Notice that the graph of $f(x)$ is composed of line segments so we can just find the endpoints of those line segments:

$$text{Figure 1: Graph of $f(x)$}$$

Step 4
4 of 7
Next, let’s determine some points on the graph of $f^{-1}(x)$ by switching the coordinates of the points for $f(x)$:

|Points on $f(x)$ |Points on $f^{-1}(x)$ |
|:–:|:–:|
|$(-2,-3)$ |$(-3,-2)$ |
|$(0,-2)$ |$(-2,0)$ |
| $(2,2)$| $(2,2)$|
| $(5,3)$| $(3,5)$|

Step 5
5 of 7
Plot the points you found for $f^{-1}(x)$ and then connect them with line segments to graph $f^{-1}(x)$:

$$text{Figure 2: Graphs of $f(x)$ and $f^{-1}(x)$}$$

Step 6
6 of 7
Lastly, let’s look at the domain and ranges of the two functions.

Recall that the domain is the set of $x$-values and the range is the set of $y$-values.

Using the graphs in Figure $2$, we can see:
* the domain of $f(x)$ is $-2le xle5$,
* the domain of $f^{-1}(x)$ is $-3le xle3$,
* the range of $f(x)$ is $-3le yle3$, and
* the range of $f^{-1}(x)$ is $-2le yle5$.

Notice that the values for the domain of $f^{-1}(x)$ are the range values of $f(x)$ and the values for the range of $f^{-1}(x)$ are the domain values of $f(x)$.

Step 7
7 of 7
**Summary:**

First, we recalled the definition of an inverse function. Next, we used the given graph to find points for $f(x)$. Applying the relationship between the points of $f(x)$ and the points of $f^{-1}(x)$, we then determined some points for the graph of $f^{-1}(x)$ using the points we found for $f(x)$. Then, we graphed $f^{-1}(x)$. Lastly, we identified the domains and ranges of the two functions and compared them.

Exercise 35
Step 1
1 of 1
Exercise scan
The wedge of cheese is part of a cylinder (approximately $dfrac{1}{8}$ of a cylinder). She can get this solid by rotating a rectangle around its width, the angle of rotation being approximately $45text{textdegree}$.
Exercise 36
Step 1
1 of 4
a)

using the calculator:

As shown in the graph below, 63% of Aurora’s

classmates are expected to be in the averageExercise scan

Step 2
2 of 4
b)

using the calculator:

As shown in the graph below, 5% of Aurora’s

classmates are expected to be in excellent shapeExercise scan

Step 3
3 of 4
c)

When the mean heart rate is 70 instead of 74:

As shown in the graph below, 20% of Aurora’s

classmates are expected to be in excellent shapeExercise scan

Result
4 of 4
a) 63 %

b) 5 %

c) 20 %

Exercise 37
Step 1
1 of 2
a) y=3(x-t)^2+2
horizontal shift right if t>0
horizontal shift left if t<0
Step 2
2 of 2
b) y=tlx-3l+4
Vertical stretch for ltl>1
Vertical compression for ltl<0
Reflection if t<0
Exercise 38
Step 1
1 of 5
a) Crystal obtain $-3$ as follows

$$
begin{align*}
sqrt {-3} cdot sqrt {-3} &= sqrt {(-3)(-3)}\
&= sqrt {(-3)^2}
end{align*}
$$

Since, the square root and square are inverse function, we obtain

$$
sqrt {-3} cdot sqrt {-3} = -3
$$

Step 2
2 of 5
b) Crystal obtain $3$ as follows

$$
begin{align*}
sqrt {-3} cdot sqrt {-3} &= sqrt {(-3)(-3)}\
&= sqrt {(-3)^2}\
&= sqrt 9\
&= 3
end{align*}
$$

Step 3
3 of 5
c) Crystal should pay attention to the following

$$
begin{align*}
sqrt {(-a)^2} &= -a\
sqrt {a^2} &= a
end{align*}
$$

Step 4
4 of 5
d) In order to apply

$$
sqrt a cdot sqrt b = sqrt {ab}
$$

it is necessary $a$ and $b$ are positive numbers.

Result
5 of 5
a) $sqrt {-3} cdot sqrt {-3} =-3$

b) $sqrt {-3} cdot sqrt {-3} =3$

c) $sqrt {(-a)^2} = -a qquad sqrt {a^2} = a$

d) Necessary $a$ and $b$ are positive numbers.

Exercise 39
Step 1
1 of 5
(a) The table describes a linear function; $y(x)=dfrac{31-(-5)}{7-1}x+c=6x+c$. Deducing $c=-11$ so that $y=6x-11$
Step 2
2 of 5
(b), (c) Interchanging $x$ and $y$ in the table gives the inverse;

Exercise scan

Step 3
3 of 5
(d) $x=6y-11$ (e) $y=dfrac{x+11}{6}$
Step 4
4 of 5
(e) Demonstrate that $y(x)=6x-11$ and $y(x)=dfrac{x+11}{6}$ are inverse functions by $y(y^{-1}(x))=x$. Here; $yleft(dfrac{x+11}{6} right)=6left( dfrac{x+11}{6}right)-11=x$
Result
5 of 5
see table and multiple solutions
Exercise 40
Step 1
1 of 2
a) $-$First find the inverse $f^{-1}$ of the given function as shown below:

$f(x)=2(x-1)^3$

$Rightarrow y=2(x-1)^3qquad$ [interchange $x$ and $y$]

$Rightarrow x=2(y-1)^3qquad$ [solve for $y$]

$Rightarrow (y-1)^3=dfrac{x}{2}qquad$ [take cube roots]

$Rightarrow sqrt[3]{(y-1)^3}=sqrt[3]{dfrac{x}{2}}qquad$

$Rightarrow y-1=sqrt[3]{dfrac{x}{2}}qquad$ [add $1$]

$Rightarrow y=sqrt[3]{dfrac{x}{2}}+1qquad$ [set $y=f^{-1}(x)$]

$Rightarrow f^{-1}(x)=sqrt[3]{dfrac{x}{2}}+1$

Step 2
2 of 2
b) For the second function follow the same steps as above:

$f(x)=3Big(dfrac{x-9}{2}Big)+20$

$Rightarrow y=3Big(dfrac{x-9}{2}Big)+20qquad$ [interchange $x$ and $y$]

$Rightarrow x=3Big(dfrac{y-9}{2}Big)+20qquad$ [solve for $y$]

$Rightarrow 3Big(dfrac{y-9}{2}Big)=x-20qquad$ [divide by $3$]

$Rightarrow dfrac{y-9}{2}=dfrac{x-20}{3}qquad$ [multiply by $2$]

$Rightarrow y-9=dfrac{2}{3}(x-20)qquad$ [add $9$]

$Rightarrow y=dfrac{2}{3}(x-20)+9qquad$ [set $y=f^{-1}(x)$]

$$
Rightarrow f^{-1}(x)=dfrac{2}{3}(x-20)+9
$$

Exercise 41
Step 1
1 of 6
$f(x)=2(x-1)^3$

$$
f^{-1}(x)=sqrt[3]{dfrac{x}{2}}+1
$$

a) We are given the functions:
Step 2
2 of 6
$x=1$

$f(1)=2(1-1)^3=0$

$f^{-1}(0)=sqrt[3]{dfrac{0}{2}}+1=1$

$Rightarrow f^{-1}(f(1))=1$

$x=2$

$f(2)=2(2-1)^3=2$

$f^{-1}(2)=sqrt[3]{dfrac{2}{2}}+1=1+1=2$

$Rightarrow f^{-1}(f(2))=2$

We choose values for $x$:
Step 3
3 of 6
$$
f(f^{-1}(x))=f(f^{-1}(x))=x
$$
b) Yes, her idea makes sense. If the functions are inverse to each other we have:
Step 4
4 of 6
c) Her strategy of choosing random values and getting a good result does not work because she cannot check this for ALL the values of $x$. She has to prove this algebraically.
Step 5
5 of 6
$f(f^{-1}(x))=fleft(sqrt[3]{dfrac{x}{2}}+1right)$

$=2left(sqrt[3]{dfrac{x}{2}}+1-1right)^3=2left(sqrt[3]{dfrac{x}{2}}right)^3$

$=2left(dfrac{x}{2}right)=x$

$f^{-1}(f(x))=f^{-1}(2(x-1)^3)$

$=sqrt[3]{dfrac{2(x-1)^3}{2}}=1=sqrt[3]{(x-1)^3}+1$

$$
=x-1+1=x
$$

d) She proves this algebraically:
Step 6
6 of 6
e) The strategy works for any input from the function’s domain.
Exercise 42
Step 1
1 of 6
$f(x)=(x-3)^2+2$

$$
f^{-1}(x)=sqrt{x-2}+3
$$

We are given the functions:
Step 2
2 of 6
$x=4$

$f(4)=(4-3)^2+2=1+2=3$

$f^{-1}(3)=sqrt{3-2}+3=1+3=4checkmark$

$x=6$

$f^{-1}(6)=sqrt{6-2}+3=2+3=5$

$f(5)=(5-3)^2+2=4+2=6checkmark$

a) We check a few values of $x$:
Step 3
3 of 6
$$
f^{-1}(f(-5))=f^{-1}((-5-3)^2+2)=f^{-1}(66)=sqrt{66-2}+3=8+3=11not=-5
$$
b) We check $f^{-1}(f(-5))$:
Step 4
4 of 6
Exercise scan
This happened because we didn’t restrict the domain of $f$ so hat the inverse is a function.
Step 5
5 of 6
$$
xin [3,infty)
$$
c) The graph of $f$ is a parabola. We restrict it to half of it:
Step 6
6 of 6
$$
f^{-1}(x)=sqrt{x-2}+3, xin [2,infty)
$$
The inverse of the restricted function is:
Exercise 43
Step 1
1 of 11
a) Three methods by which they can check the pairs of functions are actually inverses are:

– by graphing them and seeing if they are symmetrical across the $y=x$ line

– by starting with the original function and determining the formula of its inverse

– by checking that $f^{-1}(f(x))=f(f^{-1}(x))$ for any $x$ in the domain of $f$.

Step 2
2 of 11
$f(x)=dfrac{3}{5}x-15$

$$
g(x)=dfrac{5}{3}x+25
$$

b) i) We are given the functions:
Step 3
3 of 11
$f(g(x))=fleft(dfrac{5}{3}x+25right)=dfrac{3}{5}left(dfrac{5}{3}x+25right)-15$

$=x+15-15=xcheckmark$

$g(f(x))=gleft(dfrac{3}{5}x-15right)=dfrac{5}{3}left(dfrac{3}{5}x-15right)+25$

$=x-25+25=xcheckmark$

$Rightarrow f$ and $g$ are inverses

We use the third method to check if they are inverses:
Step 4
4 of 11
$j(x)=dfrac{2}{x}+10$

$$
k(x)=dfrac{2}{x-10}
$$

ii) We are given the functions:
Step 5
5 of 11
$j(k(x))=jleft(dfrac{2}{x-10}right)+10=dfrac{2}{dfrac{2}{x-10}}+10$

$=x-10+10=xcheckmark$

$k(j(x))=kleft(dfrac{2}{x}+10right)=dfrac{2}{dfrac{2}{x}+10-10}$

$=dfrac{2}{dfrac{2}{x}}=2cdot dfrac{x}{2}=xcheckmark$

$Rightarrow j$ and $k$ are inverses

We use the third method to check if they are inverses:
Step 6
6 of 11
$e(x)=dfrac{(x-10)^2}{4}$

$$
d(x)=4sqrt x+10
$$

iii) We are given the functions:
Step 7
7 of 11
$e(d(x))=e(4sqrt x+10)=dfrac{(4sqrt x+10-10)^2}{4}$

$$
=dfrac{(4sqrt x)^2}{4}=dfrac{16x}{4}=4xnot=x
$$

We use the third method to check if they are inverses:
Step 8
8 of 11
$d(x)=2sqrt x+10, xin[0,infty)$

$e(x)=dfrac{(x-10)^2}{4}, xin[10,infty)$

$e(d(x))=e(2sqrt x+10)=dfrac{(2sqrt x+10-10)^2}{4}$

$=dfrac{(2sqrt x)^2}{4}=dfrac{4x}{4}=xcheckmark$

$d(e(x))=dleft(dfrac{(x-10)^2}{4}right)+10=4sqrt{dfrac{(x-10)^2}{4}}+10$

$$
=2cdot dfrac{x-10}{2}+10=x-10+10checkmark
$$

The correct form for $d(x)$ is:
Step 9
9 of 11
$m(x)=2(x-1)^3+7$

$$
p(x)=sqrt[3]{dfrac{x-1}{2}}+7
$$

iv) We are given the functions:
Step 10
10 of 11
$m(p(x))=mleft(sqrt[3]{dfrac{x-1}{2}}+7right)$

$=2left(sqrt[3]{dfrac{x-1}{2}}+7-1right)^3+7not=x$

We use the third method to check if they are inverses:
Step 11
11 of 11
$p(x)=sqrt[3]{dfrac{x-7}{2}}+1$

$m(p(x))=mleft(sqrt[3]{dfrac{x-7}{2}}+1right)$

$=2left(sqrt[3]{dfrac{x-7}{2}}+1-1right)^3+7$

$=2left(sqrt[3]{dfrac{x-7}{2}}right)^3+7$

$=2cdot dfrac{x-7}{2}+7$

$=x-7+7=xcheckmark$

$p(m(x))=p(2(x-1)^3+7)=sqrt[3]{dfrac{2(x-1)^3+7-7}{2}}+1$

$=sqrt[3]{(x-1)^3}+1$

$$
=x-1+1=xcheckmark
$$

The correct form of $p(x)$ is:
Exercise 44
Step 1
1 of 1
To apply this exercise
1. Prepare your Learning Log.
2. Open a new page with title “Inverse Functions”.
3. Choose a function and its inverse and then justify that your functions are inverses of each other using multiple representations.
4. Explain why you have to restrict the domains of your functions.
Exercise 45
Step 1
1 of 2
Let $f(x)$ and $g(x)$ are two functions and both are inverse to each other. Then domain of $f(x)$ is range of $g(x)$, range of $f(x)$ is domain of $g(x)$, $x$ intercept of $f(x)$ is $y$ intercept of $g(x)$ and $y$ intercept of $f(x)$ is $x$ intercept of $g(x)$. So if know about the function, we can get the information of it’s inverse function
Result
2 of 2
Domain of $f(x)$ is range of $g(x)$
Exercise 46
Step 1
1 of 2
textbf{(a)} Evaluating $f(3)$:
begin{align*}
f(3)&=5(3)-3\
&=15-3\
&=12
intertext{Evaluating $g(f(3)):$}
g(f(3))&=g(12)\
&=(12-1)^2\
&=11^2\
&=121
end{align*}
textbf{(b)} Evaluating $g(3)$:
begin{align*}
g(3)&=(3-1)^2\
&=2^2\
&=4
intertext{Evaluating $f(g(3)):$}
f(g(3))&=f(4)\
&=5(4)-3\
&=20-3\
&=17\
end{align*}
Result
2 of 2
(a) $g(f(3))=121$ ~ (b) $f(g(x))=17$
Exercise 47
Step 1
1 of 14
In this exercise, we need to determine what power each number needs to be raised to get the desired result.
Step 2
2 of 14
A **power** is an expression that represents repeated multiplication of the same number:

$$a^b=underbrace{acdot acdot acdot acdots a}_{btext{ times}}$$

where $a$ is called the base and $b$ is called the exponent. The base is the number that is repeatedly multiplied and the exponent is how many times it gets multiplied.

Step 3
3 of 14
When dealing with bases that are larger than the result, the exponent will be a fraction which makes it difficult to use repeated multiplication.

In these situations, it can help to use the following exponent rules:
* **Product of Powers:** $x^acdot x^b=x^{a+b}$
* **One-to-One Property:** If $x^a=x^b$, then $a=b$.

Step 4
4 of 14
### a)
To determine the power, let’s rewrite the result of $27$ as repeated multiplication:
$$27=3cdot9=3cdot3cdot3.$$
The base of $3$ is multiplied $3$ times so the expression written using an exponent is $3^3$. Therefore, the power we had to raise $3$ to get $27$ was $boxed{3}$.
Step 5
5 of 14
### b)
To determine the power, let’s rewrite the result of $32$ as repeated multiplication:
$$32=4cdot8=(2cdot2)cdot(2cdot2cdot2).$$
The base of $2$ is multiplied $5$ times so the expression written using an exponent is $2^5$. Therefore, the power we had to raise $2$ to get $32$ was $boxed{5}$.
Step 6
6 of 14
### c)
To determine the power, let’s rewrite the result of $625$ as repeated multiplication:
$$625=25cdot25=(5cdot5)cdot(5cdot5).$$
The base of $5$ is multiplied $4$ times so the expression written using an exponent is $5^4$. Therefore, the power we had to raise $5$ to get $625$ was $boxed{4}$.
Step 7
7 of 14
### d)
The base of $64$ is larger than the desired result of $8$ so the exponent will be a fraction. Letting $n$ be the exponent we get:
$$64^n=8$$
We know that $64=8cdot8=8^2$ so we can rewrite this equation as:
$$(8^2)^n=8.$$
Using the Power of Products Property to simplify gives:
$$8^{2n}=8.$$
Since $8=8^1$ using the One-to-One Property and solving for $n$ we get:
$$begin{align*}
2n&=1\
n&=frac{1}{2}end{align*}$$
Therefore, the power we had to raise $64$ to get $8$ was $boxed{tfrac{1}{2}}$.
Step 8
8 of 14
### e)
The base of $81$ is larger than the desired result of $3$ so the exponent will be a fraction. Letting $n$ be the exponent we get:
$$81^n=3$$
We know that $81=3cdot3cdot3cdot3=3^4$ so we can rewrite this equation as:
$$(3^4)^n=3.$$
Using the Power of Products Property to simplify gives:
$$3^{4n}=3.$$
Since $3=3^1$ using the One-to-One Property and solving for $n$ we get:
$$begin{align*}
4n&=1\
n&=frac{1}{4}end{align*}$$
Therefore, the power we had to raise $81$ to get $3$ was $boxed{tfrac{1}{4}}$.
Step 9
9 of 14
### f)
The base of $64$ is larger than the desired result of $2$ so the exponent will be a fraction. Letting $n$ be the exponent we get:
$$64^n=2$$
We know that $64=2cdot2cdot2cdot2cdot2cdot2=2^6$ so we can rewrite this equation as:
$$(2^6)^n=2.$$
Using the Power of Products Property to simplify gives:
$$2^{6n}=2.$$
Since $2=2^1$ using the One-to-One Property and solving for $n$ we get:
$$begin{align*}
6n&=1\
n&=frac{1}{6}end{align*}$$
Therefore, the power we had to raise $64$ to get $2$ was $boxed{tfrac{1}{6}}$.
Step 10
10 of 14
### g)

The base of $(x^2)$ is larger than the desired result of $x^1$ so the exponent will be a fraction. Letting $n$ be the exponent we get:
$$(x^2)^n=x^1$$
Using the Power of Products Property to simplify gives:
$$x^{2n}=x^1.$$
Using the One-to-One Property and solving for $n$ we get:
$$begin{align*}
2n&=1\
n&=frac{1}{2}end{align*}$$
Therefore, the power we had to raise $(x^2)$ to get $x^1$ was $boxed{tfrac{1}{2}}$.

Step 11
11 of 14
### h)

The base of $(x^3)$ is smaller than the desired result of $x^{12}$ so we can use repeated multiplication. Let’s rewrite the result of $x^{12}$ as repeated multiplication of the factor $x^3$:
$$begin{align*}
x^{12}&=(xcdot xcdot x)cdot (xcdot xcdot x)cdot (xcdot xcdot x)cdot (xcdot xcdot x)\
&=x^3cdot x^3cdot x^3cdot x^3end{align*}$$
The base of $x^3$ is multiplied $4$ times so the expression written using an exponent is $(x^3)^4$. Therefore, the power we had to raise $(x^3)$ to get $x^{12}$ was $boxed{4}$.

Step 12
12 of 14
### i)

The power $x^a$ has a base of $x$ and an exponent of $a$. Therefore, the power we need to raise $x$ to get $x^a$ is $boxed{a}$.

Step 13
13 of 14
**Summary:**

First, we recalled the definition of a power and some of the basic exponent rules. When the desired result was larger than the base, we used repeated multiplication to find the power. When the desired result was smaller than the base, we used the Power of a Power and One-to-One Properties to find the power.

Result
14 of 14
$$begin{align*}
text{a)}&quad 3quad&text{d)}&quad tfrac{1}{2}quad&text{g)}&quadtfrac{1}{2}\
text{b)}&quad 5quad&text{e)}&quadtfrac{1}{4}quad&text{h)}&quad4\
text{c)}&quad4quad&text{f)}&quadtfrac{1}{6}quad&quadtext{i)}&quad a
end{align*}$$
Exercise 48
Step 1
1 of 5
$$
k(x)=sqrt{x+5}
$$
We are given the function:
Step 2
2 of 5
$x+5=9^2$

$$
x+5=81
$$

a) She knows that $f(x)=9$ and she needs to determine $x$. So to find $x$ she needs to determine a number which added to 5 and then to which the square root was applied leads to 9. Taking things back, she undoes the last operation:
Step 3
3 of 5
$x=81-5$

$$
x=76
$$

Then she undoes the first operation:
Step 4
4 of 5
$y=sqrt{x+5}$

$x=sqrt{y+5}$

$x^2=(sqrt{y+5})^2$

$x^2=y+5$

$y=x^2-5$

$$
k^{-1}(x)=x^2-5
$$

b) What a machine which undoes Krista’s machine is a machine which uses the inverse function of $k(x)$. We determine the inverse function:
Result
5 of 5
a) $x=76$

b) $k^{-1}(x)=x^2-5$

Exercise 49
Step 1
1 of 5
David runs the 800-meter race and Regina swims the 200-meter freestyle.
The following information is given:

Boy’s 800-meter run mean time ($bar{x}_b$) = 149 seconds.
Boy’s 800-meter run standard deviation ($sigma_b$) = 13.6 seconds.
Girl’s 200-meter freestyle mean time ($bar{x}_g$) = 145 seconds.
Girl’s 200-meter freestyle standard deviation ($sigma_g$) = 8.2 seconds.

David’s best time = 2:02 minutes or (2 $times$ 60) + 2 = 122 secs.

Regina’s best time = 2:10 minutes or (2 $times$ 60) + 10 = 130 secs.

We have to find the percentile in which each of them fall and then conclude who is a faster athlete.

Step 2
2 of 5
Here, we will use the concept of $z$- score. $z$- score tells us the distance i.e. the standard deviation between the mean and the sample score.
In this case we are given the standard deviation and the mean. The sample scores are the best times of David and Regina in their respective athletic events.

It is pertinent to note,
$$z=dfrac{x-bar{x}}{sigma}$$

Step 3
3 of 5
**David’s Case**
$$begin{aligned}
x_b&=122 \
bar{x}_b& = 149 \
sigma_b &= 13.6\
end{aligned}$$
Calculating the $z$-score,
$$begin{aligned}
z_b&=dfrac{x_b-bar{x}_b}{sigma_b} \\
&=dfrac{122-149}{13.6} \\
&=-dfrac{27}{13.6}\\
&=-1.9853
end{aligned}$$

Now we will use the normal distribution table,

We get
$z<-1.9853 = 0.023556$

This implies David falls in the $2.3556 %$

Step 4
4 of 5
**Regina’s Case**
$$begin{aligned}
x_g&=130 \
bar{x}_g& = 145 \
sigma_g &= 8.2\
end{aligned}$$
Calculating the $z$-score,
$$begin{aligned}
z_g&=dfrac{x_b-bar{x}_b}{sigma_b} \\
&=dfrac{130-145}{8.2} \\
&=-dfrac{15}{8.2}\\
&=-1.8293
end{aligned}$$

Now we will use the normal distribution table,

We get
$z<-1.8293 = 0.03368$

This implies Regina falls in the $3.368 %$

Step 5
5 of 5
Since $z_b<z_g$, i.e. David is in the 2.3556 % and Regina is in the 3.368 %, David is comparatively faster than Regina.
Exercise 50
Step 1
1 of 2
Let

$$
a+b+c = 160
$$

We know

$$
begin{align*}
a &= 56\
frac bc &= frac 58
end{align*}
$$

therefore

$$
begin{align*}
a &= 56\
b &= frac 58 c
end{align*}
$$

Now, we can calculate $c$

$$
begin{align*}
56 + frac 58 c + c &= 160\
frac {13}{8} c &= 104\
13 c &= 832\
c &= 64
end{align*}
$$

Now, we have

$$
begin{align*}
b &= frac 58 cdot 64\
b &= 5 cdot 8\
b &= 40
end{align*}
$$

The triangle sides are

$$
begin{align*}
a &= 56\
b &= 40\
c &= 64
end{align*}
$$

Result
2 of 2
$$
a=56 qquad b=40 qquad c=64
$$
Exercise 51
Step 1
1 of 1
Exercise scan
plot shown is of region bounded by inequalities
$$
begin{bmatrix}1+x-ygeq3x-2y-4\y< 2x^{2}+1end{bmatrix}
$$
The linear boundary is solid since 1+x-y can be greater than or equal to $3x-2y-4$
Exercise 52
Step 1
1 of 11
a-

i.

The graph of the function $y=x^2$ and its inverse.

The inverse of the function:

$y=x^2$

$x=sqrt {y}$

$$
y=sqrt {x}
$$

Exercise scan

Step 2
2 of 11
a-

ii.

The graph of the function $y=x^3$ and its inverse.

The inverse of the function:

$y=x^3$

$x=sqrt [3]{y}$

$$
y=sqrt [3]{x}
$$

Exercise scan

Step 3
3 of 11
a-

iii.

The graph of the function $y=x$ and its inverse.

The inverse of the function:

$y=x$

$$
y=x
$$

Exercise scan

Step 4
4 of 11
a-

iv.

The graph of the function $y=|x|$ and its inverse.

The inverse of the function:

$y=|x|$

$$
y = begin{cases}
x & xgeq 0 \
-x & xleq 0 \
end{cases}
$$

Exercise scan

Step 5
5 of 11
a-

v.

The graph of the function $y=sqrt {x}$ and its inverse.

The inverse of the function:

$y=sqrt {x}$

$y^2=x$

$$
y=x^2
$$

Exercise scan

Step 6
6 of 11
a-

vi.

The graph of the function $y=dfrac {1}{x}$ and its inverse.

The inverse of the function:

$y=dfrac {1}{x}$

$yx=1$

$$
xdfrac {1}{y}
$$

$$
y=dfrac {1}{x}
$$

Exercise scan

Step 7
7 of 11
a-

vii.

The graph of the function $y=b^x$ and its inverse.

The inverse of the function:

$y=b^x$

$log_b y=x$

$$
y=log_bx
$$

Exercise scan

Step 8
8 of 11
a-

viii.

The graph of the function $x^2+y^2=1$ and its inverse.

The inverse of the function is:

$$
x^2+y^2=1
$$

Exercise scan

Step 9
9 of 11
b-

The parent functions that are their own inverse are:

$y=x$

$y=dfrac {1}{x}$

$x^2+y^2=1$

We know because the graph of the inverse is coincident to the graph of the original function.

Step 10
10 of 11
c-

The function that fails to satisfy the horizontal line test, its inverse is not a function. They are:

$y=x^2$

$y=|x|$

$x^2+y^2=1$

Result
11 of 11
The parent functions that are their own inverse are:

$y=x$,          $y=dfrac {1}{x}$,          and          $x^2+y^2=1$

The functions that have not-function inverses are:

$y=x^2$,          $y=|x|$, and $x^2+y^2=1$

Exercise 53
Step 1
1 of 5
$$
f(x)=3^x
$$
We are given the function:
Step 2
2 of 5
begin{center}
begin{tabular}{|| c|c|c| c||}
hline
x & $f(x)$ & $(x,(f(x))$ & $(f(x),x)$ \ [0.5ex]
hline
-3 & 0.04 & (-3,0.04) & (0.04,-3) \
hline
-2 & 0.11 & (-2,0.11) & (0.11,-2)\
hline
-1 & 0.33 & (-1,0.33) & (0.33,-1)\
hline
0 & 1 & (0,1) & (1,0)\
hline
1 & 3 & (1,3) & (3,1) \
hline
2 & 9 & (3,9) & (9,3) \
hline
3 & 27 & (3,27) & (27,3) \[1ex]
hline
end{tabular}
end{center}
a) We make a table of the inverse:
Step 3
3 of 5
Exercise scan
b) We graph the inverse:
Step 4
4 of 5
$3^x=81$

$3^x=3^4$

$$
x=4
$$

c) We have to compute $f^{-1}(81)$:
Step 5
5 of 5
$$
3^y=x
$$
d) If we input $x$ for the inverse function, the output will be the exponent ($y$) of 3 which gives us $x$:
Exercise 54
Step 1
1 of 2
textbf{(a)}
begin{align*}
log_2 16&=log_2 2^4\
&=4
intertext{textbf{(b)}}
log_2 32&=log_2 2^5\
&=5
intertext{textbf{(c)}}
log_x 100&=2\
x^2&=100\
x&=10
intertext{textbf{(d)}}
log_5 x&=3\
x&=5^3\
x&=125
intertext{textbf{(e)}}
log_x 81&=4\
x^4&=9^2\
x^4&=3^4\
x&=3
intertext{textbf{(f)}}
log_{100} 10&=log_{100} sqrt{100}\
&=log_{100} 100^{1/2}\
&=dfrac{1}{2}\
end{align*}
Result
2 of 2
(a) 4 (b) 5 (c) 10 (d) 125 (e) 3 (f) $dfrac{1}{2}$
Exercise 55
Step 1
1 of 3
$$
y=3^x
$$
We are given the function:
Step 2
2 of 3
$$
log_3 y=x
$$
Using the Ancient Puzzle we can write:
Result
3 of 3
$$
log_3 y=x
$$
Exercise 56
Step 1
1 of 2
$$
y=log_3 x
$$
The Ancient Puzzle helps to find the exponent when the base and the result are given.

In Exercise 5-53 the inverse function had as input the output of $3^x$ and the base we deal with is 3.

Therefore we can write:

Result
2 of 2
$$
y=log_3 x
$$
Exercise 57
Step 1
1 of 3
a) According to the formula

$$
begin{align*}
color{#c34632}{log_ba = c} tag {*}\
color{#c34632}{a= b^c}
end{align*}
$$

we obitain

$$
begin{align*}
log_b 243 &= 5\
243 &= b^5\
b &= sqrt[5]{243}\
b &= 3
end{align*}
$$

Step 2
2 of 3
b) According to the formula $(*)$ we obtain

$$
begin{align*}
log_b 0.001 &= -3\
0.001 &= b^{-3}\
frac {1}{b^3} &= 0.001\
b^3 &= frac {1}{0.001}\
b^3 &= 1000\
b &= sqrt[3]{1000}\
b &= 10
end{align*}
$$

Result
3 of 3
a) $b = 3$

b) $b = 10$

Exercise 58
Step 1
1 of 2
Exercise scan
A cross-section of a tetrahedron can be an equilateral triangle or a square.
Step 2
2 of 2
The tetrahedron has 4 faces. When a plane cuts a face of the tetrahedron we get a side of the cross-section. Therefore the maximum number of sides a cross-section can have is 4.
Exercise 59
Step 1
1 of 4
Plotting $y(x)=2^{x}-3$ below; we have (a) domain and range $xinmathbb{R}$ and $y>-3$.

(b) There are no lines of symmetry.

Exercise scan

Step 2
2 of 4
(c) x-intercept is given by setting $y=0=2^{x}-3$ then $xln2=ln3$ giving $x=1.585$. y-intercept is given by $y(0)=-3$
Step 3
3 of 4
(d) Shifting $y=f(x)$ by k units with $kgeq3$ so that now $y=2^{x}-3+k$
Result
4 of 4
(c) $x-$intercept is $x=1.585$ and $y-$intercept is $y=-3$ (d) $y=2^x-3+k$ where $k ge 3$
Exercise 60
Step 1
1 of 2
In this exercise we can use the formula

$$
color{#c34632}{Z = frac {X-M}{s}}
$$

where

$X$ – the weight of the fruit

$M$ – the average weight of the fruit

$s$ – the standard deviation of the weights of the fruit

Apple:

$$
begin{align*}
X &= 940\
M &= 840\
s &= 120
end{align*}
$$

we have

$$
begin{align*}
Z &= frac {940 – 840}{120}\
Z &= frac {100}{120}\
Z &= 0.83
end{align*}
$$

Mango:

$$
begin{align*}
X &= 400\
M &= 350\
s &= 190
end{align*}
$$

we have

$$
begin{align*}
Z &= frac {400 – 350}{190}\
Z &= frac {50}{190}\
Z &= 0.26
end{align*}
$$

Since, $0.83>0.26$ we can conclude that new fertilizer appears to be more effective on apple trees.

Result
2 of 2
$0.83>0.26$

New fertilizer appears to be more effective on apple trees.

Exercise 61
Step 1
1 of 2
We know, equation of a circle is given by:

$$
(x-h)^2+(y-k)^2=r^2
$$
where $(h,k)$ is centre of circle

$textbf{(a)}$ Given:

$(h,k)=(-2,13)$

$r=12$

Putting given values in equation of a circle:

$$
begin{align*}
(x-(-2))^2+(y-13)^2&=(12)^2\
(x+2)^2+(y-13)^2&=144\
end{align*}
$$

$textbf{(b)}$ Given:

$(h,k)=(-1,-4)$

$r=1$

Putting given values in equation of a circle:

$$
begin{align*}
(x-(-1))^2+(y-(-4))^2&=(1)^2\
(x+1)^2+(y+4)^2&=1\
end{align*}
$$

$textbf{(c)}$ Simplifying given equation:

$$
begin{align*}
x^2+y^2-6x+16y+57&=0\
x^2-6x+9-9+y^2+16y+64-64+57&=0\
(x-3)^2-9+(y+8)^2-64+57&=0\
(x-3)^2+(y+8)^2-73+57&=0\
(x-3)^2+(y+8)^2-16&=0\
(x-3)^2+(y+8)^2&=16\
(x-3)^2+(y+8)^2&=(4)^2
end{align*}
$$

Result
2 of 2
$$
(textrm{a}) (x+2)^2+(y-13)^2=144, (textrm{b}) (x+1)^2+(y+4)^2=1, (textrm{c}) (x-3)^2+(y+8)^2=(4)^2
$$
Exercise 62
Step 1
1 of 4
a) Let

$$
begin{align*}
f(x) &= x^2 -1\
g(x) &= 3 (x+2)
end{align*}
$$

Step 2
2 of 4
We can calculate

$$
begin{align*}
g(f(x)) &= 3 ( f(x) + 2) && text{ Substitute g(x)}\
&= 3 ( ( x^2 -1) + 2) && text{ Substitute f(x)}\
&= 3 ( x^2 -1 +2) \
&= 3 (x^2 + 1) && text{Simplify}\
&= 3x^2 + 3
end{align*}
$$

Step 3
3 of 4
b) We can calculate

$$
begin{align*}
f(g(x)) &= (g(x))^2 -1 && text{ Substitute f(x)}\
&= ( 3 (x+2))^2 -1 && text{ Substitute g(x)}\
&= 9 (x+2)^2 -1\
&= 9 ( x^2 + 4x +4) -1\
&= 9x^2 + 36x + 36 -1\
&= 9x^2 + 36x +35
end{align*}
$$

Result
4 of 4
a) $g(f(x)) = 3x^2 + 3$

b) $f(g(x)) = 9x^2 + 36x +35$

Exercise 63
Step 1
1 of 6
begin{center}
begin{tabular}{|| c| c||}
hline
x & g(x) \ [0.5ex]
hline
8 & 3 \
hline
32 & 5 \
hline
$dfrac{1}{2}$ &-1 \
hline
1 & 0 \
hline
16 & 4\
hline
4 & 2 \
hline
3 & \
hline
64 & 6 \
hline
2 & 1 \
hline
0 & \
hline
0.25 & -2 \
hline
-1 & \
hline
$sqrt 2$ & \
hline
0.2 & \
hline
$dfrac{1}{8}$ & -3 \[1ex]
hline
end{tabular}
end{center}
a) We complete the table:
Step 2
2 of 6
$$
2^{g(x)}=x
$$
An equation relating $x$ and $g(x)$ is:
Step 3
3 of 6
$$
g(x)=log_2 x
$$
b) We rewrite the above equation:
Step 4
4 of 6
c) To determine an output for $x=0$ means to determine a number $y$ so that $2^y=0$. The value of $y$ should be close to $-infty$.
Step 5
5 of 6
To determine an output for $x=-1$ means to determine a number $y$ so that $2^y=-1$. As there is no number fr which an exponential is negative, the function $g$ is undefined for $xleq 0$.
Step 6
6 of 6
$y=log_2 25$

$2^y=25$

$2^{4.64}<25<2^{4.65}$

$$
yapprox 4.64
$$

d) We compute $g(25)$:
Exercise 64
Step 1
1 of 8
$$
g(x)=log_5 x
$$
We are given the function:
Step 2
2 of 8
$5^{-2}=dfrac{1}{25}$

$5^{4}=625$

a) She can fill the the values of $g(x)$ for the following values of $x$ which are of the form $5^y$:
Step 3
3 of 8
begin{center}
begin{tabular}{|| c| c||}
hline
x & g(x) \ [0.5ex]
hline
$dfrac{1}{25}$ & -2 \
hline
$dfrac{1}{5}$ & -1 \
hline
$dfrac{1}{2}$ & -0.43 \
hline
1 & 0 \
hline
2 & 0.43\
hline
3 & 0.68\
hline
4 & 0.86 \
hline
5 & 1 \
hline
6 & 1.11 \
hline
7 & 1.21 \
hline
8 & 1.29 \
hline
10 & 1.37 \
hline
25 & 2 \
hline
100 & 2.86 \
hline
125 & 3 \
hline
625 & 4 \[1ex]
hline
end{tabular}
end{center}
b) We use a calculator and properties of exponents to determine approximate values of the missing outputs:
Step 4
4 of 8
$$
5^{g(x)}=x
$$
We write an equation relating $x$ and $g(x)$:
Step 5
5 of 8
c) As $x$ increases, $g(x)$ also increases.
Step 6
6 of 8
Exercise scan
d) We plot the points from the table and sketch $g(x)$. We also sketch $f(x)=5^x$.
Step 7
7 of 8
We notice that the two functions are symmetrical across the $y=x$ line, therefore they are inverse to each other.
Result
8 of 8
(c) $g(x)$ increases as $x$ increases (d) Rotate graph of $y=log_5 x$ about $y=x$
Exercise 65
Step 1
1 of 2
Exercise scan
Step 2
2 of 2
Exercise scan
Exercise 66
Step 1
1 of 8
$$
y=log_9 x
$$
a) We are given the function:
Step 2
2 of 8
$x=log_9 y$ (switch $x$ and $y$)

$y=9^x$

We determine the inverse:
Step 3
3 of 8
$$
y=10^x
$$
b) We are given the function:
Step 4
4 of 8
$x=10^y$ (switch $x$ and $y$)

$y=log_{10} x$

We determine the inverse:
Step 5
5 of 8
$$
y=log_6 (x+1)
$$
c) We are given the function:
Step 6
6 of 8
$x=log_6 (y+1)$ (switch $x$ and $y$)

$y+1=6^x$

$$
y=6^x-1
$$

We determine the inverse:
Step 7
7 of 8
$$
y=5^{2x}
$$
d) We are given the function:
Step 8
8 of 8
$x=5^{2y}$ (switch $x$ and $y$)

$2y=log_5 x$

$$
y=0.5log_5 x
$$

We determine the inverse:
Exercise 67
Step 1
1 of 1
Exercise scan
Exercise 68
Step 1
1 of 2
$x=2^{y}$ and the original equation $y=log_{2}x$ are equivalent. By taking base 2 logarithms of both sides of the equation $x=2^{y}rightarrowlog_{2} x=log_{2}2^{y}=y$ we use the fact that if $a=b$ then $log a=log b$ to arrive back at $y=log_{2}x$
Result
2 of 2
$$
x=2^{y}
$$
Exercise 69
Step 1
1 of 2
begin {center}
renewcommand {arraystretch}{2}
begin{tabular}{|c|c|c|}
hline
& Exponential Form & Logarithmic Form \
hline
a- & $y=5^x$ & $x=log_5 (y)$ \
hline
b- & $x=7^y$ & $y=log_7 (x)$ \
hline
c- & $8^x=y$ & $x=log_8 (y)$ \
hline
d- & $A^k=C$ & $k=log_A (C)$ \
hline
e- & $A^k=C$ & $k=log_A (C)$ \
hline
f- & $K=left(dfrac {1}{2} right)^N$ & $log_{left( dfrac {1}{2} right)} (K)=N$ \
hline
end{tabular}
end {center}
Result
2 of 2
Completing the given table with the rule:

$y=a^x$ is equivalent to $x=log_a(y)$

Exercise 70
Step 1
1 of 7
$$
f(x)=dfrac{x-2}{x+2}
$$
We are given the function:
Step 2
2 of 7
begin{center}
begin{tabular}{|| c|c| c||}
hline
$x$ & $y=f(x)$ & $(x,y)$ \ [0.5ex]
hline
-20 & 1.22 & (-20,1.22) \
hline
-10 & 1.5 & (-10,1.5) \
hline
-5 & 2.33 & (-5,2.33) \
hline
-3 & 5 & (-3,5) \
hline
-2 & undefined & \
hline
-1 & -3 & (-1,-3) \
hline
0 & -1& (0,-1) \
hline
2 & 0 & (2,0) \
hline
4 & 0.33 & (4,0.33) \
hline
6 & 0.5 & (6,0.5) \
hline
8 & 0.6 & (8,0.6) \
hline
10 & 0.67 & (10,0.67) \
hline
20 & 0.82 & (20,0.82) \
hline
100 & 0.96 & (100,0.96) \[1ex]
hline
end{tabular}
end{center}
We make a table for $f(x)$:
Step 3
3 of 7
Exercise scan
We graph $f(x)$:
Step 4
4 of 7
begin{center}
begin{tabular}{|| c|c| c||}
hline
$x$ & $y=g(x)$ & $(x,y)$ \ [0.5ex]
hline
1.22 &-20 & (1.22,-20) \
hline
1.5 & -10 & (1.5,-10) \
hline
2.33 & -5 & (2.33,-5) \
hline
5 & -3 & (5,-3) \
hline
1 & undefined & \
hline
-3 & -1 & (-3,-1) \
hline
-1 & 0& (-1,0) \
hline
0 & 2 & (0,2) \
hline
0.33 & 4 & (0.33,4) \
hline
0.5 & 6 & (0.5,6) \
hline
0.6 & 8 & (0.6,8) \
hline
0.67 & 10 & (0.67,10) \
hline
0.82 & 20 & (0.82,20) \
hline
0.96 &100& (0.96,100) \[1ex]
hline
end{tabular}
end{center}
b) We make a table for the inverse $g(x)$:
Step 5
5 of 7
Exercise scan
We graph $g(x)$ (in blue color):
Step 6
6 of 7
The function $f(x)$ is undefined in $x=-2$, while $g(x)$ is undefined in $x=1$.
Step 7
7 of 7
$f(x)=dfrac{x-2}{x+2}$

$y=dfrac{x-2}{x+2}$

$x=dfrac{y-2}{y+2}$

$x(y+2)=y-2$

$xy+2x=y-2$

$xy-y=-2x-2$

$y(x-1)==2x-2$

$y=dfrac{-2x-2}{x-1}$

$g(x)=dfrac{-2x-2}{x-1}$

b) We determine the equation of $g(x)$:
Exercise 71
Step 1
1 of 8
$$
e(x)=1+sqrt{x+5}
$$
We are given the function:
Step 2
2 of 8
$x+5geq 0Rightarrow xgeq -5$

The domain: $[-5,infty)$

a) We determine the domain of $e(x)$:
Step 3
3 of 8
$y=1+sqrt{x+5}$

$x=1+sqrt{y+5}$

$x-1=sqrt{y+5}$

$(x-1)^2=(sqrt{y+5})^2$

$y+5=(x-1)^2$

$y=(x-1)^2-5$

$$
e^{-1}(x)=(x-1)^2-5
$$

b) We determine $e^{-1}(x)$:
Step 4
4 of 8
$$
[1,infty)
$$
The domain of $e^{-1}(x)$ is:
Step 5
5 of 8
$e^{-1}(e(-4))=e^{-1}(1+sqrt{-4+5})$

$$
=e^{-1}(2)=(2-1)^2-5=-4
$$

c) We determine $e^{-1}(e(-4))$:
Step 6
6 of 8
We got $e^{-1}(e(x))=x$ because the two functions are inverse t each other.
Step 7
7 of 8
d) The two graphs are symmetrical across the line $y=x$.
Step 8
8 of 8
Exercise scan
e) We draw both graphs:
Exercise 72
Step 1
1 of 3
The parent graph of the function

$$
f(x) =-2 (x-1)^2 + 3
$$

Exercise scan

Step 2
2 of 3
We can transform a given function in the following way

$$
f(x) =-2 (x-1)^2 +3
$$

$3$ units down:

$$
f(x) = -2 (x-2)^2
$$

$1$ unit left:

$$
f(x) = -2 x^2
$$

Exercise scan

Result
3 of 3
$$
f(x)=-2(x-1)^2+3
$$

$$
f(x)=-2x^2
$$

Exercise 73
Step 1
1 of 3
The shaded area (the overlapping area) is $dfrac{1}{4}$ from the area of one of the squares.

Let’s note:

$A$=the area of one of the two congruent squares.

Step 2
2 of 3
$P(text{not in the shaded area})=dfrac{dfrac{3}{4}A+dfrac{3}{4}A}{2A-dfrac{1}{4}A}$

$=dfrac{dfrac{6}{4}A}{dfrac{7}{4}A}$

$=dfrac{6}{4}cdotdfrac{4}{7}$

$$
=dfrac{6}{7}
$$

We determine the probability that the point is not in the shaded area:
Result
3 of 3
$$
dfrac{6}{7}
$$
Exercise 74
Step 1
1 of 2
$2^{(x+y)}=16$          (Given)

$2^{(x+y)}=2^4$

$x+y=4$

$y=4-x$ qquad (1)

$2^{(2x+y)}=dfrac {1}{8}$          (Given)

$2^{(2x+y)}=2^{-3}$

$2x+y=-3$

$y=-2x-3$          (2)

$4-x=-2x-3$          (Equating (1) and (2))

$2x-x=-3-4$          (Grouping similar terms)

$$
x=-7
$$

Substituting for $x$ in equation (1)

$y=4-(-7)$

$$
y=11
$$

Result
2 of 2
The solution is the point $(-7, 11)$
Exercise 75
Step 1
1 of 6
**Parent Function:**
$y=2^x$

The parent function, $y=2^x$, when translated to a logarithmic form, generates $x=log_{2}y$.

It will also generate the following table:

|$x$|$y$|
|–|–|
|-5|$dfrac{1}{32}$|
|-4|$dfrac{1}{16}$|
|-3|$dfrac{1}{8}$|
|-2|$dfrac{1}{4}$|
|-1|$dfrac{1}{2}$|
|0|0|
|1|2|
|2|4|
|3|8|
|4|16|
|5|32|

Step 2
2 of 6
**How many tables are required?**
As there is an infinite possibility of the set of values of the domain, there is also an infinite number of table of values to be written.
Domain: $reals$ or $(-infty, infty)$
Step 3
3 of 6
**Relationship with logarithms and inverses and the function’s graph:**
When graphing the function, $y=2^x$, it is important to note that the input value of $x$ will result to a certain value, $y$ and the input value of $-x$ will result to a certain value, $dfrac{1}{y}$.
![‘slader’](https://slader-solution-uploads.s3.amazonaws.com/49cae4c7-3611-4b79-aec5-1fbda02ab2c9-1632721137004275.png)
Step 4
4 of 6
**Patterns in the Graph:**
Positive values of $x$ will render values of $y$ that increase exponentially. Negative values of $x$ will render values of $y$ that decrease gradually.
Step 5
5 of 6
**Possible Inputs and Outputs and Different Representations:**
As mentioned, there are inifinite values of the domain as it tend to go to $infty$ from $-infty$.

$text{D: } reals, (-infty,infty)$
$space$
All $x$-values make sense as positive $x$-values exponentially increase and negative $x$-values gradually decrease.
$space$
![‘slader’](https://slader-solution-uploads.s3.amazonaws.com/b0a7fc86-1d8c-42df-ad51-54c77cf92f0f-1632723282697727.png)
This is also true to any base. However, it can be seen that the higher the base the closer the graph tends to be drawn in to the $y$-axis.

Step 6
6 of 6
**When $b$ is altered:**
In any positive $b$-value, the concept that positive $x$-values exponentially increase and negative $x$-values gradually decrease is true.

However, for a negative $b$-value, like $y=(-2)^x$, the function’s graph seems discontinuous when plotted.
![‘slader’](https://slader-solution-uploads.s3.amazonaws.com/193a32b9-98c1-4435-bf91-595e6bff6849-1632723747094842.png)

Exercise 76
Step 1
1 of 2
Let $b=2$ , then the table of $y=log_2{x}$ is shown below

$$
begin{array}{|c|c|}
hline x&y\ hline
1&0\
hline
2&1\ hline
4&2\
hline8&3\
hline16&4\
hline32&5\
hline
end{array}
$$
then the inverse of $y=log_2{x}$ is
$$
x=2^y
$$

$$
begin{array}{|c|c|}
hline y&x\ hline
1&0\
hline
2&1\ hline
4&2\
hline8&3\
hline16&4\
hline32&5\
hline
end{array}
$$
From the calculator also we will get the same values

Let $y=2^x$ then it is same as
$$
y=log_2{x}
$$

Result
2 of 2
$$
y=log_2{x}
$$
Exercise 77
Step 1
1 of 1
To apply this exercise
1. Prepare your Learning Log.
2. Open a new page with title “The Family of Logarithmic Functions “.
3. Write the descriptive statements your team has come up with.
4. Write a list of statements are very clear to you.
5. Write a list of statements are need further clarification.
Exercise 78
Step 1
1 of 2
Noting $log a=log b$ if $a=b$ then we can take base 7 logarithms of both sides of the expression; $log_{7}x=log_{7}(7^{y})=y$. So we use the fact that $log_{a}a^{b}=b$ for $a>0$
Result
2 of 2
$$
y=log_{7}x
$$
Exercise 79
Step 1
1 of 2
The rule that applies to logarithms is

$$
color{#c34632}{log_aa = 1}
$$

According to the formula

$$
color{#c34632}{ log {a^b} = b log {a}}
$$

we obtain

$$
log_6{6^{11}} = 11 log_6{6} = 11 cdot 1 = 11
$$

Therefore

$$
log_6{6^{11}} = 11
$$

Result
2 of 2
$$
log_6{6^{11}} = 11
$$
Exercise 80
Step 1
1 of 3
Exercise scan
A vertical cross-section would lead to 2 separate circles:
Step 2
2 of 3
Exercise scan
A horizontal cross-section would lead to a ring:
Result
3 of 3
2 circles; a ring
Exercise 81
Step 1
1 of 11
$$
f(x)=dfrac{1}{x-3}+4
$$
a) We are given the function:
Step 2
2 of 11
Exercise scan
The parent function is $y=dfrac{1}{x}$. We shift the parent function 3 units to the right to get $y=dfrac{1}{x-3}$, then shift the result 4 units up to get $f(x)$:
Step 3
3 of 11
$$
h(x)=3sqrt x
$$
b) We are given the function:
Step 4
4 of 11
Exercise scan
We start with parent function $y=sqrt x$ which we vertically stretch by a factor of 3 to get $h(x)$:
Step 5
5 of 11
$(h,k)=(0,-3)$

$h=0$

$k=-3$

$$
y=ax^2-3
$$

c) The parabola has the vertex:
Step 6
6 of 11
$5=a(2^2)-3$

$5=4a-3$

$4a=8$

$$
a=2
$$

We determine $a$ from the function’s equation:

$y=a(x-h)^2+k$,

using the point $(2,5)$:

Step 7
7 of 11
$$
y=2x^2-3
$$
The function is:
Step 8
8 of 11
$$
y=a(x-h)^3+k
$$
d) The function has the form:
Step 9
9 of 11
$h=1$

$k=4$

$$
Rightarrow y=a(x-1)^3+4
$$

We determine $h, k$:
Step 10
10 of 11
$5=a(0-1)^3+4$

$5=-a+4$

$a=4-5$

$$
a=-1
$$

We determine $a$ using the point $(0,5)$:
Step 11
11 of 11
$$
y=-(x-1)^3+4
$$
The function is:
Exercise 82
Step 1
1 of 7
National Health Statistics Report states that average height of an adult woman in the U.S. is 63.8 inches. The standard deviation is 2.7 inches.

$$bar x = 63.8$$
$$sigma = 2.7$$

Step 2
2 of 7
a. We have to find the percentage of women in the US that are under 4 feet 11 inches tall.

4 feet 11 inches implies 4 $times$12+11=59 inches.

Here, we will use the concept of $z$- score. $z$- score tells us the distance i.e. the standard deviation between the mean and the sample score.
In this case we are given the standard deviation and the mean. The sample score is the percentage of women in the US that are under 4 feet 11 inches tall.

It is pertinent to note,
$$z=dfrac{x-bar{x}}{sigma}$$

Step 3
3 of 7
Now, calculating the $z$-score,
$$begin{aligned}
z_a&=dfrac{x-bar{x}}{sigma} \\
&=dfrac{59-63.8}{2.7} \\
&=-dfrac{4.8}{2.7}\\
&=-1.7778
end{aligned}$$

Now we will use the normal distribution table,

We get
$z<-1.7778 = 0.03772$

This implies $3.772 %$ of women in the US are under 4 feet 11 inches.

Step 4
4 of 7
b. Now, it is given that most girls reach their adult height by their high school senior year.

In the North City Senior High School there are 324 students of whom 50 % are assumed to be girls.

Thus number of girl students = $dfrac{50}{100} times 324 = 162$

Fro part a. above we have concluded that 3.772 % of women in the US are under 4 feet 11 inches.

Therefore, 3.772 % of girl students in North City High School are shorter than the height of 4 feet 11 inches

Calculating,

$dfrac{3.772}{100} times 162 = 6.11$

Thus, around 6 girls in the North City High School are are shorter than the height of 4 feet 11 inches.

Step 5
5 of 7
c. We have to predict the percentage of women in the North City High School that are taller than 6 feet.

6 feet implies 6 $times$ 12 =72 inches.

now first we will find the percentage of women in the US that are taller than 6 feet. Here also, we will use the concept of $z$- score.

Now, calculating the $z$-score,
$$begin{aligned}
z_c&=dfrac{x-bar{x}}{sigma} \\
&=dfrac{72-63.8}{2.7} \\
&=-dfrac{9.2}{2.7}\\
&=3.4074
end{aligned}$$

Now we will use the normal distribution table,

We get
$z>3.4074 = 0.001195$

This implies $0.1195 %$ of women in the US are taller than 6 feet.

Step 6
6 of 7
Accordingly on the assumption that the distribution of women in the US is somewhat similar in the North City High School, it can be implied that 0.1195 % of the girl students in the school (162) are taller than 6 feet.

Calculating,

$dfrac{0.1195}{100} times 162 = 0.194$

Thus, no girls in the North City High School are are taller than the height of 6 feet.

And the assumption is that an outlier (if any) is not present in the North City High School.

Step 7
7 of 7
d. On the basis of the assumption that the school in the question is a private boarding school with a championship girl’s basketball program, and they happen to recruit basketball players from across the state and the necessary implication that usually 6 feet or taller candidates are considered for the game because of the game’s very nature, it can be said that the answer in part c. (no girls in the North City High School are are taller than the height of 6 feet) will surely change.
Exercise 83
Step 1
1 of 2
$textbf{a).}$

$$
begin{align*}
x^3&=243 tag{Given}\ \
sqrt[3]{x^3}&=sqrt[3]{243} tag{Root cubing on both side}\ \
x&=sqrt[3]{3times3timestimes3times3} tag{Expanding 243}\ \
x&=3cdot sqrt[3]{9} tag{Cube of 3 is 27}\ \
x&=3times2.08 tag{$sqrt[3]{9}=2.08$}\ \
x&=6.24 tag{Simplifying}\
end{align*}
$$

$textbf{b).}$

$$
begin{align*}
3^x&=243 tag{Given}\ \
log_{3}3^x&=log_{3}243tag{Taking log on both side}\ \
xlog_{3}3&=log_{3}(27times9) tag{$log_{a}b^c=clog_{a}b$}\ \
x&=log_{3}(3^3+3^2) tag{$log_{a}a=1$}\ \
x&=log_{3}3^3+log_{3}3^2 tag{Using log addition property}\ \
x&=3log_{3}3+2log_{3}2 tag{simplifying}\ \
x&=3times1+2times1 tag{$log_{a}a=1$}\ \
x&=5tag{Simplifying}
end{align*}
$$

Result
2 of 2
$$
a.)=6.24 & b.)=5
$$
Exercise 84
Step 1
1 of 9
In this exercise, we are given two different functions and need to evaluate them at specific values of $x$, determine the value of $x$ if the functions equal a given value, and find the inverse of one of the functions.
Step 2
2 of 9
To **evaluate a function** for a specific value, we simply replace each variable in the function with the value. For example, $f(2)$ means substitute $x=2$ into the function $f(x)$ and then simplify.

To **find the inverse of a function**, perform the following steps:
* Replace $f(x)$ with $y$.
* Switch $x$ and $y$.
* Solve the equation for $y$.
* Replace $y$ with $f^{-1}(x)$.

Step 3
3 of 9
### a)
To find the value of $f(-7)$ substitute $x=-7$ into the given function and then simplify:
$$begin{align*}
f(x)&=-2x^2-4\
f(-7)&=-2(7)^2-4\
&=-2(49)-4\
&=-98-4\
&=-102.
end{align*}$$
Therefore, $f(-7)=-102$.
Step 4
4 of 9
### b)
To find the value of $g(-w)$ substitute $x=-w$ into the given function and then simplify:
$$begin{align*}
g(x)&=5x+3\
g(-2)&=5(-2)+3\
&=-10+3\
&=-7.
end{align*}$$
Therefore, $g(-2)=-7$.
Step 5
5 of 9
### c)

To find the value of $x$ such that $f(x)=c$, replace $f(x)$ in the given function with $c$ and then solve for $x$:
$$begin{align*}
f(x)&=-2x^2-4\
c&=-2x^2-4\
c+4&=-2x^2\
frac{c+4}{-2}&=x^2\
pmsqrt{frac{c+4}{-2}}&=x.
end{align*}$$
Therefore, if $f(x)=c$, then $x=sqrt{frac{c+4}{-2}}$ or $x=-sqrt{frac{c+4}{-2}}$.

Note: Remember when square rooting both sides of an equation you must write $pm$.

Step 6
6 of 9
### d)

To find the value of $x$ such that $g(x)=c$, replace $g(x)$ in the given function with $c$ and then solve for $x$:
$$begin{align*}
g(x)&=5x+3\
c&=5x+3\
c-3&=5x\
frac{c-3}{5}&=x.
end{align*}$$
Therefore, if $g(x)=c$, then $x=frac{c-3}{5}$.

Step 7
7 of 9
### e)

Following the steps for finding an inverse, we first need to replace $g(x)$ with $y$:
$$begin{align*}
g(x)&=5x+3\
y&=5x+3.
end{align*}$$
Switching $x$ and $y$ and then solving for $y$ gives:
$$begin{align*}
x&=5y+3\
x-3&=5y\
frac{x-3}{5}&=y.
end{align*}$$
Replacing $y$ with $g^{-1}(x)$, we get $g^{-1}(x)=frac{x-3}{5}$.

Step 8
8 of 9
**Summary:**

First, we recalled how to evaluate a function and the steps to find the inverse of a function. Then we evaluated $f(x)$ and $g(x)$ at the two given values of $x$. Then we solved for $x$ when each function equaled $c$. Lastly, we found the inverse function for $g(x)$.

Result
9 of 9
$$begin{align*}
text{a)}&quad f(-7)=-102quad&text{d)}&quad x=tfrac{c-3}{5}\
text{b)}&quad g(-2)=-7&text{e)}&quad g^{-1}(x)=tfrac{x-3}{5}\
text{c)}&quad x=pmsqrt{tfrac{c+4}{-2}}quad
end{align*}$$
Exercise 85
Step 1
1 of 4
$$
y=log (x)
$$
We are given the function:
Step 2
2 of 4
begin{center}
begin{tabular}{|| c| c||}
hline
$x$ & $log(x)$ \ [0.5ex]
hline
0.1 & -1 \
hline
0.5 &-0.30103 \
hline
1 &0 \
hline
2 & 0.30103 \
hline
5 & 0.69897 \
hline
10 & 1 \
hline
20 &1.30103 \
hline
50 & 1.69897 \
hline
100 & 2 \
hline
200 &2.30103 \
hline
500 & 2.69897 \
hline
1000 & 3 \
hline
2000 &3.30103 \
hline
5000 & 3.69897 \
hline
10000 & 4 \[1ex]
hline
end{tabular}
end{center}
Using our calculator, we make a table of values:
Step 3
3 of 4
$a^1=10$

$$
a=10
$$

As $log_a (10)=1$, we have:
Step 4
4 of 4
$$
a^y=x
$$
The base for $log (x)$ is 10.

The input values which give us whole number outputs are the number in the form $a^y$, where $a$ is the base.

We can rewrite $log_a (x)=y$ in an exponential form:

Exercise 86
Step 1
1 of 8
begin{center}
begin{tabular}{|| c| c||}
hline
x & f(x) \ [0.5ex]
hline
$textcolor{blue}{0.000001}$ & -6 \
hline
$textcolor{blue}{0.00001}$ & -5 \
hline
$textcolor{blue}{0.0001}$ & -4 \
hline
$textcolor{blue}{0.001}$ & -3 \
hline
$textcolor{blue}{0.01}$ & -2 \
hline
$textcolor{blue}{0.1}$ & -1 \
hline
$textcolor{blue}{1}$ & 0 \
hline
1 & $textcolor{blue}{0}$ \
hline
2 & $textcolor{blue}{0.30103}$ \
hline
3 & $textcolor{blue}{0.47712}$ \
hline
4 & $textcolor{blue}{0.60206}$ \
hline
5 & $textcolor{blue}{0.69897}$ \
hline
6 & $textcolor{blue}{0.77815}$ \[1ex]
hline
end{tabular}
end{center}
a) We complete the table for $f(x)=log(x)$:
Step 2
2 of 8
$$
x=0
$$
b) The asymptote is:
Step 3
3 of 8
Exercise scan
We graph the function, using the points in the table:
Step 4
4 of 8
$g(x)=alog (x)$

Exercise scan

c) A transformation of the graph of $f(x)=log (x)$ is vertically stretching/shrinking by a factor of $a$:
Step 5
5 of 8
$g(x)=log (bx)$

Exercise scan

A transformation of the graph of $f(x)=log (x)$ is a horizontal shrink/stretch by a factor of $b$:
Step 6
6 of 8
$g(x)=log (x+c)$

Exercise scan

A transformation of the graph of $f(x)=log (x)$ is a horizontal shift by $c$ units:
Step 7
7 of 8
$g(x)=log (x)+d$

Exercise scan

A transformation of the graph of $f(x)=log (x)$ is a vertical shift by $d$ units:
Step 8
8 of 8
$$
g(x)=alog (bx+c)+d
$$
The general form for the family of logarithmic functions is:
Exercise 87
Step 1
1 of 8
$$
f_1(x)=log(x)+3
$$
a) We are given the function:
Step 2
2 of 8
Exercise scan
We graph the parent function $f(x)=log (x)$, then we shift it 3 units up to get $f_1(x)$:
Step 3
3 of 8
$$
f(x)=log(x-2)
$$
b) We are given the function:
Step 4
4 of 8
Exercise scan
We graph the parent function $f(x)=log (x)$, then we shift it 2 units to the right to get $f_2(x)$:
Step 5
5 of 8
$$
f_3(x)=4log(x+3)-2
$$
c) We are given the function:
Step 6
6 of 8
Exercise scan
We graph the parent function $f(x)=log (x)$, then we shift it 3 units to the left to get $f_5(x)=log (x+3)$, then we vertically stretch $f_5$ by a factor of 2 to get $f_6(x)=4log(x+3)$, then we shift $f_6$ 2 units down to get $f_3(x)$:
Step 7
7 of 8
$f_4(x)=log_2(x)+3$

$f_4(x)=dfrac{log (x)}{log 2}+3$

$f_4(x)=dfrac{1}{log 2}cdot log(x)+3$

$$
f_4(x)=3.32log (x)+3
$$

d) We are given the function:
Step 8
8 of 8
Exercise scan
We graph the parent function $f(x)=log (x)$, then we vertically stretch it by a factor of 3.32 to get $f_7(x)=3.32log(x)$, then we shift it 3 units up to get $f_4(x)$:
Exercise 88
Step 1
1 of 4
a-

The calculator assumes the default base of the log is 10 and it says $log (6) approx 0.778$ because $10^{0.778}=6$

Step 2
2 of 4
b-

The base of $log (6)$ is 10.

$log (6)$ has a value less than 1 but greater than 0.

because $10^0=16$

Step 3
3 of 4
c-

A logarithm is the exponent needed for a base number to become another number.

for example $log_2(32)=5$ because the exponent needed for 2 to become 32 is 5.

Logarithmic equation is the inverse of the exponential equation.

$log_2(32)=5$ is the inverse of $2^5=32$

Result
4 of 4
a-          $log (6) approx 0.778$ because $10^{0.778} approx 6$

b-          $log (6)$ has a value less than 1 but greater than 0. because $10^0=16$

c-          A logarithm is the exponent needed for a base number to become another number.

Logarithmic equation is the inverse of the exponential equation.

Exercise 89
Step 1
1 of 7
(a) $25=x^{1}$ ; $x=25$
Solve for x by taking logarithms$;$
Step 2
2 of 7
(b) $9=3^{x}$ ; $x=2$
Step 3
3 of 7
(c) $x=7^{3}$ ; $x=343$
Step 4
4 of 7
(d) $x=sqrt{3}$
Step 5
5 of 7
(e) $27=x^{3}$ ; $x=3$
Step 6
6 of 7
(f) $10000=10^{x}$ ; $x=4$
Result
7 of 7
see answers
Exercise 90
Step 1
1 of 2
The graph of the function

$$
y = 3 log {(x+4)}
$$

Exercise scan

Result
2 of 2
$$
y = 3 log {(x+4)}
$$
Exercise 91
Step 1
1 of 1
If we let $x=log_{3}(2)$ then $2=3^{x}$ and then we let $y=log_{2}(3)$ such that $3=2^{y}$. For x we solve by taking log to base 10 ; $xlog(3)=log(2)$ ; $x=0.63$. For y we solve similarly ; $ylog(2)=log(3)$ ; $y=1.58$. Therefore $xne y$
Exercise 92
Step 1
1 of 3
(a) Consider natural numbers $aepsilonmathbb{N}$ and since we know that if the exponent $10$. Writing $log_{a}10=0.926628408$ then plugging values for $a$ into the expression $a^{0.926628408}=10$ gives a close solution when $a=12$ so for the aliens, $log12=1$
Step 2
2 of 3
(b) The aliens must have 12 fingers$.$
Result
3 of 3
base 12
Exercise 93
Step 1
1 of 1
Exercise scan
Rotating a parabola around the line $m$ leads to a solid similar to a Bundt cake form.
Exercise 94
Step 1
1 of 1
The domain of any logarithmic function is all positive real numbers $Rtextgreater0$. Since y=24 is in the domain we can take the logarithm of y=24=k where k is a constant and get a constant solution.
Exercise 95
Step 1
1 of 5
$f(x)=sqrt{7-x}-6$

$$
f^{-1}(x)=-(x+6)^2+7
$$

We are given the functions:
Step 2
2 of 5
Domain: $(-infty,7]$

Range: $[-6,infty)$

a) We determine the domain and range of $f(x)$:
Step 3
3 of 5
Domain: $[-6,infty)$

Range: $(-infty,7]$

We determine the domain and range of $f^{-1}(x)$, restricted so that it is the inverse of $f(x)$:
Step 4
4 of 5
$$
f^{-1}(f(a))=a
$$
b) We predict:
Step 5
5 of 5
$f^{-1}(f(a))=f^{-1}(sqrt{7-a}-6)$

$=-(sqrt{7-a}-6+6)^2+7$

$=-(sqrt{7-a})^2+7$

$=-7+a+7$

$$
=acheckmark
$$

We check the prediction algebraically:
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