All Solutions
Page 225: Questions
$10 cdot (x+4)=-70$ (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)
$$
x=-11
$$
x=-11
$$
The following table shows the input and output of the machine.\
begin {center}
begin{tabular}{|r|r|}
hline
Input & Output \
hline
3 & 7 \
hline
4 & 9 \
hline
-3 & -5 \
hline
end{tabular}
end {center}
The machine multiplies the input by 2 and add 1 to the sum to generate the result.\
If the machine is reversing its function, the expected outputs for the different inputs will be.
For pulling back 7 the value that comes out is 3.
For pulling back 9 the value that comes out is 4.
For pulling back -5 the value that comes out is -3.
The table below demonstrates the inputs and outputs of the backwards function.\
begin {center}
begin{tabular}{|r|r|}
hline
x & y \
hline
7 & 3 \
hline
9 & 4 \
hline
-5 & -3 \
hline
end{tabular}
end {center}
Anita’s backward function is subtracting from the input and dividing the difference by 2.\
Anita’s original function machine is:
$y=2x+1$
Anita’s backward function machine is:
$y=dfrac {x-1}{2}$
b- The expected outputs for the different inputs of the reverse function will be.
3 for input 7, 4 for input 9 and -3 for input -5
c- Anita’s backward function is subtracting from the input and dividing the difference by 2.
d- Anita’s original function machine is: $y=2x+1$
Anita’s backward function machine is: $y=dfrac {x-1}{2}$
$h(4)=5 cdot 4+2=22$
22 is the number that should be pulled up into the machine in order to have a 4 come out.
Keiko’s machine must subtract 2 from 17 and divide the difference by 5 to undo the original function.
The inverse function is:
$h^{-1}(x)=dfrac {x-2}{5}$
Let’s check the original function $h(x)$ and the inverse function $h^{-1}(x)$ for $x=4$
$h(4)=5 cdot 4+2=22$
$h^{-1}(22)=dfrac {22-2}{5}=dfrac {20}{5}=4$
b- Keiko’s machine must subtract 2 from 17 and divide the difference by 5 to undo the original function.
c- The inverse function is: $h^{-1}(x)=dfrac {x-2}{5}$
d- $h(4)=5 cdot 4+2=22$
$h^{-1}(22)=dfrac {22-2}{5}=dfrac {20}{5}=4$
$$
g^{-1}(x)=sqrt[3]{x+5}
$$
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}
$color{#c34632} text{$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:
$y=dfrac{1}{2}x-3qquadqquadqquadqquadqquad$ $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:
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*}
*How do we determine the shape of a cross-section?*
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.
$$text{Figure 1: Sketch of a couch}$$
Note: Couches can vary in size and shape so your sketch may have some differences compared to Figure $1$.
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.
$$
begin{align*}
1^x &= 5\
ln 1^x &= ln 5
end{align*}
$$
Now we can use the formula
$$
begin{align*}
color{#c34632}{ln a^b = b ln a} tag {*}
end{align*}
$$
$$
begin{align*}
x ln 1 &= ln 5\
x &= frac {ln 5}{ln 1}\
x &= frac {1.61}{0} &&text{( $ln 1=0$)}\
x &= infty
end{align*}
$$
The solution of this equation does not exist.
$$
begin{align*}
sqrt {27^x} &= 81\
27^x &= 81^2\
27^x &= 6561
end{align*}
$$
Now we can use the $ln$ function
$$
ln 27^x = ln 6561
$$
$$
begin{align*}
x ln 27 &= ln 6561\
x &= frac {ln 6561}{ln 27}\
x &= frac {8.78}{3.29}\
x &= 2.66
end{align*}
$$
The solution of this equation is
$$
x = 2.66
$$
$$
begin{align*}
2^x &= 9\
ln 2^x &= ln 9
end{align*}
$$
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
$$
x = 3.17
$$
$$
begin{align*}
25^{(x+1)} &= 125^x\
(5^2)^{(x+1)} &= (5^3)^x \
5^{2(x+1)} &= 5^{3x} && text{( $(a^n)^m=a^{n cdot m}$ )}
end{align*}
$$
$$
begin{align*}
color{#c34632}{a^b = a^c} tag {**}\
color{#c34632}{b=c}
end{align*}
$$
we obtain
$$
begin{align*}
2( x+1 ) &= 3x\
2x + 2 &= 3x\
2 &= 3x-2x\
x &= 2
end{align*}
$$
The solution of this equation is
$$
x=2
$$
$$
begin{align*}
8^x &= 2^5 cdot 4^4\
left( 2^3right)^x &= 2^5 left( 2^2 right)^4\
2^{3x} &= 2^5 cdot 2^8
end{align*}
$$
$$
color{#c34632}{a^b cdot a^c = a^{b+c}}
$$
Now we have
$$
begin{align*}
2^{3x} &= 2^{5+8}\
2^{3x} &= 2^{13}
end{align*}
$$
$$
begin{align*}
3x &= 13\
x &= frac {13}{3}\
x &= 4.33
end{align*}
$$
The solution of this equation is
$$
x = 4.33
$$
b) $x = 2.66$
c) $x = 3.17$
d) $x = 2$
e) $x = 4.33$
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
$$
y = 3 cdot 2^x
$$
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
$$
y = 10 cdot 2^x
$$
$$
begin{align*}
y = 3 cdot 2^x\
y = 10 cdot 2^x
end{align*}
$$
b) $y = 10 cdot 2^x$
c) Graph function $y=10 cdot 2^x$ grows faster than graph function $y= 3 cdot 2^x$.
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.
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.
$$
AREA=text{length}times text{width}=3sqrt{2}times sqrt{2}=3times 2=6
$$
$$
11.5 ( 1.083)^t
$$
can mean that the price of ticket rose by $1.083$ times.
The ticket is growing every month for $8.3 %$.
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}
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}
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}
y=0.5x+3
$$
y=3(x+2)^2-6
$$
y=dfrac{1}{6}x^3-dfrac{13}{6}x+2
$$
y=x
$$
y=left(dfrac{x}{2}right)^2
$$
Because the inverse parabola is not a function, we must work with restrictions: for example the original function restricted to $(0,infty)$ has the inverse the upper half of the inverse curve.
$$
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 )
$$
$$
y = left( frac x2 right)^2
$$
is rank of the function
$$
x = pm 2 sqrt y
$$
and vice versa.
$R = ( – infty, + infty )$
b) The domain of the function
$$
y = left( frac x2 right)^2
$$
is rank of the function
$$
x = pm 2 sqrt y
$$
and vice versa.
y=left(dfrac{x}{2}right)^2
$$
y=pm2sqrt{x}
$$
g(x)=(x-3)^2
$$
[3,infty)
$$
$y=(x-3)^2$
$x=(y-3)^2$
$sqrt x=y-3$
$y=sqrt x+3$
$$
g^{-1}(x)=sqrt x+3, xgeq 0
$$
According to the horizontal line test, any horizontal line must intersect the graph of $f$ at most in one point.
If there is more than one points of intersection the inverse of $f$ will not be a function.
An example of a function $f$ whose inverse is not a function is the function:
$f(x)=x^2$
The inverse of this function is not a function because the graph of $f$ does not pass the horizontal line test.
Sketch the graph of the inverse function $f^{-1}(x)$(green colour) with symmetry with respect to the line $y=x$.
See the picture below:
$y=1+dfrac{2}{x}qquad$ [interchange $x$ and $y$]
$Rightarrow x=1+dfrac{2}{y}qquad$ [silve for $y$]
$Rightarrow x-1=dfrac{2}{y}qquad$ [cross multiplication]
$Rightarrow y=dfrac{2}{x-1}qquad$ [set $y=f^{-1}(x)$]
$$
Rightarrow f^{-1}(x)=dfrac{2}{x-1}
$$
$$
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)
$$
$f(x)=dfrac{1}{2}(x-1)^3$
$Rightarrow y=dfrac{1}{2}(x-1)^3qquad$ [interchange $x$ and $y$]
$Rightarrow x=dfrac{1}{2}(y-1)^3qquad$ [solve for $y$]
$Rightarrow (y-1)^3=2xqquad$ [take cube roots]
$Rightarrow y-1=sqrt[3]{2x}qquad$ [add $1$]
$Rightarrow y=sqrt[3]{2x}+1qquad$ [set $y=f^{-1}(x)$]
$Rightarrow f^{-1}(x)=sqrt[3]{2x}+1$
$-$The graphs of $f$(red colour) and $f^{-1}$(blue colour) is shown in the pic below.
f^{-1}=sqrt[3]{2x}+1
$$
$y=3x-8$ (Given)
To find the inverse function:
$y+8=3x$ (Add 8 to each side)
$x=dfrac {y+8}{3}$
The inverse function is:
$$
y=dfrac {x+8}{3}
$$
$y=dfrac {1}{2}x+6$ (Given)
To find the inverse function:
$2y=x+12$ (Multiply each side by 2)
$x=2y-12$
The inverse function is:
$$
y=2x-12
$$
$y=dfrac {x+6}{2}$ (Given)
$2y=x+6$ (Multiply each side by 2)
$x=2y-6$
The inverse function is:
$$
y=2x-6
$$
b- $y=2x-12$
c- $y=2x-6$
* 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.
* **Quotient of Powers Property:**
$frac{x^a}{x^b}=x^{a-b}$
* **One-to-One Property:** If $x^a=x^b$, then $a=b$.
Since $b^0=1$ for all $bne0$, we can simplify the first equation to get the value of $a$:
$$begin{align*}
3&=acdot b^0\
3&=acdot1\
3&=a.
end{align*}$$
$$begin{align*}
75&=acdot b^2\
75&=3cdot b^2\
frac{75}{3}&=frac{3cdot b^2}{3}\
25&=b^2\
5^2&=b^2\
5&=b.
end{align*}$$
Solve the first equation for $a$:
$$begin{align*}
18&=acdot b^2\
frac{18}{b^2}&=frac{acdot b^2}{b^2}\
frac{18}{b^2}&=a.
end{align*}$$
$$begin{align*}
54&=acdot b^3\
54&=frac{18}{b^2}cdot b^3\
54&=18cdot b^{3-2}\
54&=18b\
frac{54}{18}&=frac{18b}{18}\
3&=b.
end{align*}$$
$$begin{align*}
18&=acdot b^2\
18&=acdot 3^2\
18&=acdot9\
frac{18}{9}&=frac{acdot 9}{9}\
2&=a.
end{align*}$$
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.
b) $a=2$ and $b=3$
$$L(x)=x^2-1$$
$$R(x)=3(x+2)$$
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.\\
$$(x+2)^2+(y+2)^2=(2r)^2$$
Add 2 to the x side and subtract 5 to the y side. Look at solution below.
$$
begin{bmatrix}xgeq-3\xleq3\yleq-dfrac{2}{3}x+3\ygeq-dfrac{2}{3}x-3end{bmatrix}
$$
$f(x)=5(x-2)$
$Rightarrow y=5(x-2)qquad$ [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.
*When given a point for a function $f(x)$, how can we find a point for $f^{-1}(x)$?*
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 $f^{-1}(x)$.
$$text{Figure 1: Graph of $f(x)$}$$
|Points on $f(x)$ |Points on $f^{-1}(x)$ |
|:–:|:–:|
|$(-2,-3)$ |$(-3,-2)$ |
|$(0,-2)$ |$(-2,0)$ |
| $(2,2)$| $(2,2)$|
| $(5,3)$| $(3,5)$|
$$text{Figure 2: Graphs of $f(x)$ and $f^{-1}(x)$}$$
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 $f(x)$.
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.
using the calculator:
As shown in the graph below, 63% of Aurora’s
classmates are expected to be in the average
using the calculator:
As shown in the graph below, 5% of Aurora’s
classmates are expected to be in excellent shape
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 shape
b) 5 %
c) 20 %
horizontal shift right if t>0
horizontal shift left if t<0
Vertical stretch for ltl>1
Vertical compression for ltl<0
Reflection if t<0
$$
begin{align*}
sqrt {-3} cdot sqrt {-3} &= sqrt {(-3)(-3)}\
&= sqrt {(-3)^2}
end{align*}
$$
Since, the square root and square are inverse function, we obtain
$$
sqrt {-3} cdot sqrt {-3} = -3
$$
$$
begin{align*}
sqrt {-3} cdot sqrt {-3} &= sqrt {(-3)(-3)}\
&= sqrt {(-3)^2}\
&= sqrt 9\
&= 3
end{align*}
$$
$$
begin{align*}
sqrt {(-a)^2} &= -a\
sqrt {a^2} &= a
end{align*}
$$
$$
sqrt a cdot sqrt b = sqrt {ab}
$$
it is necessary $a$ and $b$ are positive numbers.
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.
$f(x)=2(x-1)^3$
$Rightarrow y=2(x-1)^3qquad$ [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 $y=f^{-1}(x)$]
$Rightarrow f^{-1}(x)=sqrt[3]{dfrac{x}{2}}+1$
$f(x)=3Big(dfrac{x-9}{2}Big)+20$
$Rightarrow y=3Big(dfrac{x-9}{2}Big)+20qquad$ [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 $y=f^{-1}(x)$]
$$
Rightarrow f^{-1}(x)=dfrac{2}{3}(x-20)+9
$$
$$
f^{-1}(x)=sqrt[3]{dfrac{x}{2}}+1
$$
$f(1)=2(1-1)^3=0$
$f^{-1}(0)=sqrt[3]{dfrac{0}{2}}+1=1$
$Rightarrow f^{-1}(f(1))=1$
$x=2$
$f(2)=2(2-1)^3=2$
$f^{-1}(2)=sqrt[3]{dfrac{2}{2}}+1=1+1=2$
$Rightarrow f^{-1}(f(2))=2$
f(f^{-1}(x))=f(f^{-1}(x))=x
$$
$=2left(sqrt[3]{dfrac{x}{2}}+1-1right)^3=2left(sqrt[3]{dfrac{x}{2}}right)^3$
$=2left(dfrac{x}{2}right)=x$
$f^{-1}(f(x))=f^{-1}(2(x-1)^3)$
$=sqrt[3]{dfrac{2(x-1)^3}{2}}=1=sqrt[3]{(x-1)^3}+1$
$$
=x-1+1=x
$$
$$
f^{-1}(x)=sqrt{x-2}+3
$$
$f(4)=(4-3)^2+2=1+2=3$
$f^{-1}(3)=sqrt{3-2}+3=1+3=4checkmark$
$x=6$
$f^{-1}(6)=sqrt{6-2}+3=2+3=5$
$f(5)=(5-3)^2+2=4+2=6checkmark$
f^{-1}(f(-5))=f^{-1}((-5-3)^2+2)=f^{-1}(66)=sqrt{66-2}+3=8+3=11not=-5
$$
xin [3,infty)
$$
f^{-1}(x)=sqrt{x-2}+3, xin [2,infty)
$$
– 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 $f$.
$$
g(x)=dfrac{5}{3}x+25
$$
$=x+15-15=xcheckmark$
$g(f(x))=gleft(dfrac{3}{5}x-15right)=dfrac{5}{3}left(dfrac{3}{5}x-15right)+25$
$=x-25+25=xcheckmark$
$Rightarrow f$ and $g$ are inverses
$$
k(x)=dfrac{2}{x-10}
$$
$=x-10+10=xcheckmark$
$k(j(x))=kleft(dfrac{2}{x}+10right)=dfrac{2}{dfrac{2}{x}+10-10}$
$=dfrac{2}{dfrac{2}{x}}=2cdot dfrac{x}{2}=xcheckmark$
$Rightarrow j$ and $k$ are inverses
$$
d(x)=4sqrt x+10
$$
$$
=dfrac{(4sqrt x)^2}{4}=dfrac{16x}{4}=4xnot=x
$$
$e(x)=dfrac{(x-10)^2}{4}, xin[10,infty)$
$e(d(x))=e(2sqrt x+10)=dfrac{(2sqrt x+10-10)^2}{4}$
$=dfrac{(2sqrt x)^2}{4}=dfrac{4x}{4}=xcheckmark$
$d(e(x))=dleft(dfrac{(x-10)^2}{4}right)+10=4sqrt{dfrac{(x-10)^2}{4}}+10$
$$
=2cdot dfrac{x-10}{2}+10=x-10+10checkmark
$$
$$
p(x)=sqrt[3]{dfrac{x-1}{2}}+7
$$
$=2left(sqrt[3]{dfrac{x-1}{2}}+7-1right)^3+7not=x$
$m(p(x))=mleft(sqrt[3]{dfrac{x-7}{2}}+1right)$
$=2left(sqrt[3]{dfrac{x-7}{2}}+1-1right)^3+7$
$=2left(sqrt[3]{dfrac{x-7}{2}}right)^3+7$
$=2cdot dfrac{x-7}{2}+7$
$=x-7+7=xcheckmark$
$p(m(x))=p(2(x-1)^3+7)=sqrt[3]{dfrac{2(x-1)^3+7-7}{2}}+1$
$=sqrt[3]{(x-1)^3}+1$
$$
=x-1+1=xcheckmark
$$
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.
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*}
$$a^b=underbrace{acdot acdot acdot acdots a}_{btext{ times}}$$
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.
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 $a=b$.
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 $boxed{3}$.
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 $boxed{5}$.
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 $boxed{4}$.
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 $boxed{tfrac{1}{2}}$.
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 $boxed{tfrac{1}{4}}$.
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 $boxed{tfrac{1}{6}}$.
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 $boxed{tfrac{1}{2}}$.
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 $boxed{4}$.
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 $boxed{a}$.
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.
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*}$$
k(x)=sqrt{x+5}
$$
$$
x+5=81
$$
$$
x=76
$$
$x=sqrt{y+5}$
$x^2=(sqrt{y+5})^2$
$x^2=y+5$
$y=x^2-5$
$$
k^{-1}(x)=x^2-5
$$
b) $k^{-1}(x)=x^2-5$
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.
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,
$$z=dfrac{x-bar{x}}{sigma}$$
$$begin{aligned}
x_b&=122 \
bar{x}_b& = 149 \
sigma_b &= 13.6\
end{aligned}$$
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 $2.3556 %$
$$begin{aligned}
x_g&=130 \
bar{x}_g& = 145 \
sigma_g &= 8.2\
end{aligned}$$
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 $3.368 %$
$$
a+b+c = 160
$$
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
$$
begin{align*}
a &= 56\
b &= 40\
c &= 64
end{align*}
$$
a=56 qquad b=40 qquad c=64
$$
$$
begin{bmatrix}1+x-ygeq3x-2y-4\y< 2x^{2}+1end{bmatrix}
$$
The linear boundary is solid since 1+x-y can be greater than or equal to $3x-2y-4$
i.
The graph of the function $y=x^2$ and its inverse.
The inverse of the function:
$y=x^2$
$x=sqrt {y}$
$$
y=sqrt {x}
$$
ii.
The graph of the function $y=x^3$ and its inverse.
The inverse of the function:
$y=x^3$
$x=sqrt [3]{y}$
$$
y=sqrt [3]{x}
$$
iii.
The graph of the function $y=x$ and its inverse.
The inverse of the function:
$y=x$
$$
y=x
$$
iv.
The graph of the function $y=|x|$ and its inverse.
The inverse of the function:
$y=|x|$
$$
y = begin{cases}
x & xgeq 0 \
-x & xleq 0 \
end{cases}
$$
v.
The graph of the function $y=sqrt {x}$ and its inverse.
The inverse of the function:
$y=sqrt {x}$
$y^2=x$
$$
y=x^2
$$
vi.
The graph of the function $y=dfrac {1}{x}$ and its inverse.
The inverse of the function:
$y=dfrac {1}{x}$
$yx=1$
$$
xdfrac {1}{y}
$$
$$
y=dfrac {1}{x}
$$
vii.
The graph of the function $y=b^x$ and its inverse.
The inverse of the function:
$y=b^x$
$log_b y=x$
$$
y=log_bx
$$
viii.
The graph of the function $x^2+y^2=1$ and its inverse.
The inverse of the function is:
$$
x^2+y^2=1
$$
The parent functions that are their own inverse are:
$y=x$
$y=dfrac {1}{x}$
$x^2+y^2=1$
We know because the graph of the inverse is coincident to the graph of the original function.
The function that fails to satisfy the horizontal line test, its inverse is not a function. They are:
$y=x^2$
$y=|x|$
$x^2+y^2=1$
$y=x$, $y=dfrac {1}{x}$, and $x^2+y^2=1$
The functions that have not-function inverses are:
$y=x^2$, $y=|x|$, and $x^2+y^2=1$
f(x)=3^x
$$
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}
$3^x=3^4$
$$
x=4
$$
3^y=x
$$
begin{align*}
log_2 16&=log_2 2^4\
&=4
intertext{textbf{(b)}}
log_2 32&=log_2 2^5\
&=5
intertext{textbf{(c)}}
log_x 100&=2\
x^2&=100\
x&=10
intertext{textbf{(d)}}
log_5 x&=3\
x&=5^3\
x&=125
intertext{textbf{(e)}}
log_x 81&=4\
x^4&=9^2\
x^4&=3^4\
x&=3
intertext{textbf{(f)}}
log_{100} 10&=log_{100} sqrt{100}\
&=log_{100} 100^{1/2}\
&=dfrac{1}{2}\
end{align*}
y=3^x
$$
log_3 y=x
$$
log_3 y=x
$$
y=log_3 x
$$
In Exercise 5-53 the inverse function had as input the output of $3^x$ and the base we deal with is 3.
Therefore we can write:
y=log_3 x
$$
$$
begin{align*}
color{#c34632}{log_ba = c} tag {*}\
color{#c34632}{a= b^c}
end{align*}
$$
we obitain
$$
begin{align*}
log_b 243 &= 5\
243 &= b^5\
b &= sqrt[5]{243}\
b &= 3
end{align*}
$$
$$
begin{align*}
log_b 0.001 &= -3\
0.001 &= b^{-3}\
frac {1}{b^3} &= 0.001\
b^3 &= frac {1}{0.001}\
b^3 &= 1000\
b &= sqrt[3]{1000}\
b &= 10
end{align*}
$$
b) $b = 10$
(b) There are no lines of symmetry.
$$
color{#c34632}{Z = frac {X-M}{s}}
$$
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.
New fertilizer appears to be more effective on apple trees.
$$
(x-h)^2+(y-k)^2=r^2
$$
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:
$$
begin{align*}
x^2+y^2-6x+16y+57&=0\
x^2-6x+9-9+y^2+16y+64-64+57&=0\
(x-3)^2-9+(y+8)^2-64+57&=0\
(x-3)^2+(y+8)^2-73+57&=0\
(x-3)^2+(y+8)^2-16&=0\
(x-3)^2+(y+8)^2&=16\
(x-3)^2+(y+8)^2&=(4)^2
end{align*}
$$
(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
$$
$$
begin{align*}
f(x) &= x^2 -1\
g(x) &= 3 (x+2)
end{align*}
$$
$$
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*}
$$
$$
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*}
$$
b) $f(g(x)) = 9x^2 + 36x +35$
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}
2^{g(x)}=x
$$
g(x)=log_2 x
$$
$2^y=25$
$2^{4.64}<25<2^{4.65}$
$$
yapprox 4.64
$$
g(x)=log_5 x
$$
$5^{4}=625$
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}
5^{g(x)}=x
$$
y=log_9 x
$$
$y=9^x$
y=10^x
$$
$y=log_{10} x$
y=log_6 (x+1)
$$
$y+1=6^x$
$$
y=6^x-1
$$
y=5^{2x}
$$
$2y=log_5 x$
$$
y=0.5log_5 x
$$
x=2^{y}
$$
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}
$y=a^x$ is equivalent to $x=log_a(y)$
f(x)=dfrac{x-2}{x+2}
$$
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}
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}
$y=dfrac{x-2}{x+2}$
$x=dfrac{y-2}{y+2}$
$x(y+2)=y-2$
$xy+2x=y-2$
$xy-y=-2x-2$
$y(x-1)==2x-2$
$y=dfrac{-2x-2}{x-1}$
$g(x)=dfrac{-2x-2}{x-1}$
e(x)=1+sqrt{x+5}
$$
The domain: $[-5,infty)$
$x=1+sqrt{y+5}$
$x-1=sqrt{y+5}$
$(x-1)^2=(sqrt{y+5})^2$
$y+5=(x-1)^2$
$y=(x-1)^2-5$
$$
e^{-1}(x)=(x-1)^2-5
$$
[1,infty)
$$
$$
=e^{-1}(2)=(2-1)^2-5=-4
$$
$$
f(x) =-2 (x-1)^2 + 3
$$
$$
f(x) =-2 (x-1)^2 +3
$$
$3$ units down:
$$
f(x) = -2 (x-2)^2
$$
$1$ unit left:
$$
f(x) = -2 x^2
$$
f(x)=-2(x-1)^2+3
$$
$$
f(x)=-2x^2
$$
Let’s note:
$A$=the area of one of the two congruent squares.
$=dfrac{dfrac{6}{4}A}{dfrac{7}{4}A}$
$=dfrac{6}{4}cdotdfrac{4}{7}$
$$
=dfrac{6}{7}
$$
dfrac{6}{7}
$$
$2^{(x+y)}=2^4$
$x+y=4$
$y=4-x$ 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)
$y=4-(-7)$
$$
y=11
$$
$y=2^x$
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|
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 $(-infty, infty)$
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)
Positive values of $x$ will render values of $y$ that increase exponentially. Negative values of $x$ will render values of $y$ that decrease gradually.
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.
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)
$$
begin{array}{|c|c|}
hline x&y\ hline
1&0\
hline
2&1\ hline
4&2\
hline8&3\
hline16&4\
hline32&5\
hline
end{array}
$$
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
$$
y=log_2{x}
$$
y=log_2{x}
$$
1. Prepare your Learning Log.
2. Open a new page with title “The Family of Logarithmic Functions “.
3. Write the descriptive statements your team has come up with.
4. Write a list of statements are very clear to you.
5. Write a list of statements are need further clarification.
y=log_{7}x
$$
$$
color{#c34632}{log_aa = 1}
$$
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
$$
log_6{6^{11}} = 11
$$
log_6{6^{11}} = 11
$$
f(x)=dfrac{1}{x-3}+4
$$
h(x)=3sqrt x
$$
$h=0$
$k=-3$
$$
y=ax^2-3
$$
$5=4a-3$
$4a=8$
$$
a=2
$$
$y=a(x-h)^2+k$,
using the point $(2,5)$:
y=2x^2-3
$$
y=a(x-h)^3+k
$$
$k=4$
$$
Rightarrow y=a(x-1)^3+4
$$
$5=-a+4$
$a=4-5$
$$
a=-1
$$
y=-(x-1)^3+4
$$
$$bar x = 63.8$$
$$sigma = 2.7$$
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,
$$z=dfrac{x-bar{x}}{sigma}$$
$$begin{aligned}
z_a&=dfrac{x-bar{x}}{sigma} \\
&=dfrac{59-63.8}{2.7} \\
&=-dfrac{4.8}{2.7}\\
&=-1.7778
end{aligned}$$
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.
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.
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.
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.
$$
begin{align*}
x^3&=243 tag{Given}\ \
sqrt[3]{x^3}&=sqrt[3]{243} tag{Root cubing on both side}\ \
x&=sqrt[3]{3times3timestimes3times3} tag{Expanding 243}\ \
x&=3cdot sqrt[3]{9} tag{Cube of 3 is 27}\ \
x&=3times2.08 tag{$sqrt[3]{9}=2.08$}\ \
x&=6.24 tag{Simplifying}\
end{align*}
$$
$textbf{b).}$
$$
begin{align*}
3^x&=243 tag{Given}\ \
log_{3}3^x&=log_{3}243tag{Taking log on both side}\ \
xlog_{3}3&=log_{3}(27times9) tag{$log_{a}b^c=clog_{a}b$}\ \
x&=log_{3}(3^3+3^2) tag{$log_{a}a=1$}\ \
x&=log_{3}3^3+log_{3}3^2 tag{Using log addition property}\ \
x&=3log_{3}3+2log_{3}2 tag{simplifying}\ \
x&=3times1+2times1 tag{$log_{a}a=1$}\ \
x&=5tag{Simplifying}
end{align*}
$$
a.)=6.24 & b.)=5
$$
To **find the inverse of a function**, perform the following steps:
* Replace $f(x)$ with $y$.
* Switch $x$ and $y$.
* Solve the equation for $y$.
* Replace $y$ with $f^{-1}(x)$.
To find the value of $f(-7)$ substitute $x=-7$ into the given function and then simplify:
$$begin{align*}
f(x)&=-2x^2-4\
f(-7)&=-2(7)^2-4\
&=-2(49)-4\
&=-98-4\
&=-102.
end{align*}$$
Therefore, $f(-7)=-102$.
To find the value of $g(-w)$ substitute $x=-w$ into the given function and then simplify:
$$begin{align*}
g(x)&=5x+3\
g(-2)&=5(-2)+3\
&=-10+3\
&=-7.
end{align*}$$
Therefore, $g(-2)=-7$.
To find the value of $x$ such that $f(x)=c$, replace $f(x)$ in the given function with $c$ and then solve for $x$:
$$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*}$$
Therefore, if $f(x)=c$, then $x=sqrt{frac{c+4}{-2}}$ or $x=-sqrt{frac{c+4}{-2}}$.
Note: Remember when square rooting both sides of an equation you must write $pm$.
To find the value of $x$ such that $g(x)=c$, replace $g(x)$ in the given function with $c$ and then solve for $x$:
$$begin{align*}
g(x)&=5x+3\
c&=5x+3\
c-3&=5x\
frac{c-3}{5}&=x.
end{align*}$$
Therefore, if $g(x)=c$, then $x=frac{c-3}{5}$.
Following the steps for finding an inverse, we first need to replace $g(x)$ with $y$:
$$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*}$$
Replacing $y$ with $g^{-1}(x)$, we get $g^{-1}(x)=frac{x-3}{5}$.
First, we recalled how to evaluate a function and the steps to find the inverse of a function. Then we evaluated $f(x)$ and $g(x)$ at the two given values of $x$. Then we solved for $x$ when each function equaled $c$. Lastly, we found the inverse function for $g(x)$.
text{a)}&quad f(-7)=-102quad&text{d)}&quad x=tfrac{c-3}{5}\
text{b)}&quad g(-2)=-7&text{e)}&quad g^{-1}(x)=tfrac{x-3}{5}\
text{c)}&quad x=pmsqrt{tfrac{c+4}{-2}}quad
end{align*}$$
y=log (x)
$$
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}
$$
a=10
$$
a^y=x
$$
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:
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}
x=0
$$
g(x)=alog (bx+c)+d
$$
f_1(x)=log(x)+3
$$
f(x)=log(x-2)
$$
f_3(x)=4log(x+3)-2
$$
$f_4(x)=dfrac{log (x)}{log 2}+3$
$f_4(x)=dfrac{1}{log 2}cdot log(x)+3$
$$
f_4(x)=3.32log (x)+3
$$
The calculator assumes the default base of the log is 10 and it says $log (6) approx 0.778$ because $10^{0.778}=6$
The base of $log (6)$ is 10.
$log (6)$ has a value less than 1 but greater than 0.
because $10^0=16$
A logarithm is the exponent needed for a base number to become another number.
for example $log_2(32)=5$ because the exponent needed for 2 to become 32 is 5.
Logarithmic equation is the inverse of the exponential equation.
$log_2(32)=5$ is the inverse of $2^5=32$
b- $log (6)$ has a value less than 1 but greater than 0. because $10^0=16$
c- A logarithm is the exponent needed for a base number to become another number.
Logarithmic equation is the inverse of the exponential equation.
$$
y = 3 log {(x+4)}
$$
y = 3 log {(x+4)}
$$
$$
f^{-1}(x)=-(x+6)^2+7
$$
Range: $[-6,infty)$
Range: $(-infty,7]$
f^{-1}(f(a))=a
$$
$=-(sqrt{7-a}-6+6)^2+7$
$=-(sqrt{7-a})^2+7$
$=-7+a+7$
$$
=acheckmark
$$