Solve the following equation by factoring.

$$
color{#4257b2}text{(a)} x^4-16x^2+75=2x^2-6
$$

Isolate the variables on the left side as follows:

$$
x^4-16x^2-2x^2=-6-75 x^4-18x^2=-81
$$

If the equation factorable, then the $(f(1))$ or $(f(-1))$ should be equal zero as follows:

$$
f(1)=1-18=-17 f(-1)=1-18=-17
$$

Since the $f(1)$ and $f(-1)$ is not equal right side, so the polynomial of this equation is not factored

$$
color{#4257b2}text{(b)} 2x^2+4x-1=x+1
$$

Isolate the variables on the left side as follows:

$$
2x^2+4x-x=1+1 2x^2+3x=2
$$

If the equation factorable, then the $(f(1))$ or $(f(-1))$ should be equal zero as follows:

$$
f(1)=2+3=5 f(-1)=2-3=-1
$$

Since the $f(1)$ and $f(-1)$ is not equal right side, so the polynomial of this equation is not factored

$$
color{#4257b2}text{(c)} 4x^3-x^2-2x+2=3x^3-2(x^2-1)
$$

Isolate the variables on the left side as follows:

$$
4x^3-x^2-2x+2=3x^3-2x^2+2
$$

$$
4x^3-x^2-2x-3x^3+2x^2=2-2=0 x^3+x^2-2x=0
$$

If the equation factorable, then the $(f(1))$ or $(f(-1))$ should be equal zero as follows:

$$
f(1)=1+1-2=0
$$

Since the $f(1)$ is equal right side, so the polynomial of this equation is factored

$$
x(x^2+x-2)=0 x(x-1)(x+2)=0
$$

Use zero property as follows:

$$
x=0 x=1 x=-2
$$

$$
color{#4257b2}text{(d)} -2x^2+x-6=-x^3+2x-8
$$

Isolate the variables on the left side as follows:

$$
-2x^2+x^3-2x+x-6+8=0 x^3-2x^2-x+2=0
$$

If the equation factorable, then the $(f(1))$ should be equal zero as follows:

$$
f(1)=1-2-1+2=0
$$

Since the $f(1)$ is equal right side, so the polynomial of this equation is factored

$$
x^2(x-2)-x+2=0 (x-2)(x^2-1)=0
$$

Use zero property as follows:

$$
x=2 x=1 x=-1
$$

$$
text{color{Brown}(a) Since the $f(1)$ and $f(-1)$ is not equal right side, so the polynomial of this equation is not factored
\ \
(b) Since the $f(1)$ and $f(-1)$ is not equal right side, so the polynomial of this equation is not factored
\ \
(c) $x=0, x=1 x=-2$ (d) $x=-1, x=1, x=2$}
$$

We would like to solve the following expression algebraically.

$$
color{#4257b2}18x^4-53x^3+52x^2-14x-8=3x^4-x^3+2x-8
$$

Use zero property to simplify as follows:

$$
18x^4-53x^3+52x^2-14x-8-3x^4+x^3-2x+8=0
$$

Rearrange the tiles to group like terms as follows:

$$
(18x^4-3x^4)+(-53x^3+x^3)+52x^2+(-14x-2x)+(-8+8)=0
$$

$$
15x^4-52x^3+52x^2-16x=0
$$

Substitute difference value of $(x)$ to know the expression is factored or not as follows:

$$
f(1)=15-52+52-16 f(1)=-1
$$

$$
f(2)=240-416+208-32 f(2)=0
$$

The term of $(x+2)$ is factored of the expression, so we use the theorem property to get other zeros as follows:

$$
2| 15 -52 52 -16
$$

$$
30 -44 16
$$

$$
15 -22 8 0
$$

$$
15x^4-52x^3+52x^2-16x=(x-2)(15x^3-22×62+8x)
$$

$$
15x^4-52x^3+52x^2-16x=x(x-2)(15x^2-22x+8)
$$

$$
15x^4-52x^3+52x^2-16x=x(x-2)(5x-4)(3x-2)
$$

$$
text{color{Brown}$x(x-2)(5x-4)(3x-2)$}
$$

$$
text{color{#4257b2}(a) Write the equation of polynomial of $f(x)$ that has a degree of $4$ and zeros, $(x=1, 2, -2, -1)$ and has $y$ intercept is $4$}
$$

According the rule of standard form for quartic equation as follows:

$$
f(x)=a(x+x_{1})+(x+x_{2})+(x+x_{3})+(x+x_{4})+C
$$

$$
f(x)=a(x-1)+(x-2)+(x+2)+(x+1)+4
$$

$f(x)=4$, when $x=0$, so:

$$
4=a[-1cdot-2cdot2cdot1] 4=4a a=1
$$

$$
f(x)=(x-1)+(x-2)+(x+2)+(x+1)+4
$$

$$
text{color{#4257b2}(b) Write the values of $x$ when $f(x)=48$}
$$

$$
48=(x-1)+(x-2)+(x+2)+(x+1)+4
$$

$$
44=(x-1)+(x-2)+(x+2)+(x+1)
$$

Use zero property as follows:

$$
x-1=44 x=45
$$

$$
x-2=44 x=46
$$

$$
x+2=44 x=42
$$

$$
x+1=44 x=43
$$

$$
text{color{Brown}(a) $f(x)=(x-1)+(x-2)+(x+2)+(x+1)+4$
\ \
(b) $x=45 x=46 x=42 x=43$}
$$

$$
text{color{#4257b2} An open topped box made from rectangle piece of cardboard with dimension of $(24)$ cm by $(30)$cm by cutting congruent squares from each corner. Calculate the dimensions of the box with a volume of $(1040)$ cm$^3$}
$$

Volume of new box can be represented by the following expression:

$$
V(x)=x(30-2x)(24-2x)
$$

Use distributive property as follows:

$$
1040=x left[720-60x-48x+4x^2right] V(x)=4x^3-108x^2+720x=1040
$$

$$
{V}'(x)=12x^2-216x+720=0
$$

Divide the entire equation by $(12)$ as follows:

$$
x^2-18x+60=0
$$

Use discriminant property for quadratic equation $left[x=dfrac{-bpmsqrt{b^2-4ac}}{2a}right]$ as follows:

$$
a=1 b=-18 c=60
$$

$$
x=dfrac{18pmsqrt{(-18)^2-(4cdot1cdot60)}}{2cdot1} x=dfrac{18pmsqrt{324-240}}{2}
$$

$$
x=dfrac{18+sqrt{84}}{2} x=13.582
$$

$$
x=dfrac{18-sqrt{84}}{2} x=4.417
$$

$$
V(x)=x(30-2x)(24-2x)
$$

Use zero property to find the range value of $(x)$ as follows:

$$
30-2xge0 -2xge-30 xle15
$$

$$
24-2xge0 -2xge-24 xle12
$$

According these ranges values of $(x)$, the matched value of $(x)$ is equal, $(4.417)$

The dimensions of box are:

Height $=4.417$ cm

Length $=30-2(4.417)=21.166$ cm

Width $=24-2(4.417)=15.166$ cm

$$
text{color{Brown}Height $=4.417$ cm
\ \
Length $=30-2(4.417)=21.166$ cm
\ \
Width $=24-2(4.417)=15.166$ cm}
$$

Between $(1985)$ through $(1959)$ the number of home computer in thousand estimated by $C(t)=0.92[t^3+8t^2+40t+400]$, where $(t=0)$ at year of $(1985)$. Answer for the following questions.

$$
text{color{#4257b2}(a) Explain why you can predict the number of computer solid by usig this modelled in year $(1993)$m but cant be allowable in year $(2005)$?}
$$

Because this expression is valid for a certain years from $(1985)$ to year of $(1995)$, and this year is out of the range.

$$
text{color{#4257b2}(b) Explain how to find when the number of computer solid in Canada reaches to $(1.5)$ million, using this model.}
$$

We substitute the value of $[C(t)=1.5]$ million, then calculate the variable time $(t)$ to get the which year the number of computer solid is reached to $(1.5)$ million.

$$
text{color{#4257b2}(c) In what year did home compute sales reach to $(1.5)$ million.}
$$

Substitute the value of $(C(t)=1.5)$ million as follows:

$$
1,500,000=0.92[t^3+8t^2+40t+400]
$$

Divide both of sides by $(0.92)$ as follows:

$$
1630434.783=t^3+8t^2+40t+400 t^3+8t^2+40t=1630034.782
$$

$$
{C}'(t)=3t^2+16t+40=0
$$

Use discriminant value $(sqrt{b^2-4ac})$ to know if the function has a real roots or not as follows:

$$
a=3 b=16 c=40
$$

$$
=sqrt{(16)^2-[4cdot3cdot40]} =sqrt{256-480}=sqrt{-224}
$$

This function has a no solution, due to there are no real roots for a negative number under square roots.

This modelled is not valid to predict which year the number of computer solid reached to $(1.5)$ million, because there are no real root for this equation at this value of $C(t)$.

That mean this value of computer solid may be reached before or after this period of years that expression is valid for it.

$$
text{color{Brown}(a) Because this expression is valid for certain years from $(1985)$ to the year of $(1995)$, and this year is out of the range.
\ \
(b) We substitute the value of $[C(t)=1.5]$ million, then calculate the variable time $(t)$ to get the which year the number of computers solid is reached to $(1.5)$ million.
\ \
(c) This function has a no solution, due to there are no real roots for a negative number under square roots.}
$$

Create a linear inequality for the following solution.

$$
color{#4257b2}text{(a)} x>8
$$

$$
2(x+2)>x+12
$$

$$
color{#4257b2}text{(b)} xge-4
$$

$$
3x+8ge2x+4
$$

$$
color{#4257b2}text{(c)} xle-12
$$

$$
2x+7ge3x-5
$$

$$
color{#4257b2}text{(d)} -9<x<-1
$$

$$
2x-4<3x+5<2x+4
$$

$$
text{color{Brown}(a) $2(x+2)>x+12$ (b) $3x+8ge2x+4$
\ \
(c) $2x+7ge3x-5$ (d) $2x-4<3x+5<2x+4$}
$$

Solve the following inequalities and set your solution in notation.

$$
color{#4257b2}text{(a)} 2(4x-7)>4(x+9)
$$

Use distributive property as follows:

$$
8x-14>4x+36
$$

Isolate the variables on the left side as follows:

$$
8x-4x>36+14 4x>50
$$

Divide entire inequality by $(4)$ as follows:

$$
x>12.5 therefore text{The solution set is} (12.5, infty)
$$

$$
color{#4257b2}text{(b)} dfrac{x-4}{5}gedfrac{2x+3}{2}
$$

Multiply entire inequality by $(10)$ as follows:

$$
2(x-4)ge5(2x+3)
$$

Use distributive property as follows:

$$
2x-8ge10x+15
$$

Isolate the variables on the left side as follows:

$$
2x-10xge15+8 -8xge23
$$

Divide entire inequality by $(-8)$ as follows:

$color{#4257b2}diamond Note that: Dividing by a negative number reverses the inequality sign.$

$$
xledfrac{-23}{8} therefore text{The solution set is} left(-infty, -dfrac{23}{8}right]
$$

$$
color{#4257b2}text{(c)} -x+2>x-2
$$

Isolate the variables on the left side as follows:

$$
-x-x>-2-2 -2x>-4
$$

Divide entire inequality by $(-2)$ as follows:

$color{#4257b2}diamond Note that: Dividing by a negative number reverses the inequality sign.$

$$
x<2 therefore text{The solution set is} (-infty, 2)
$$

$$
color{#4257b2}text{(d)} 5x-7le2x+2
$$

Isolate the variables on the left side as follows:

$$
5x-2xle2+7 3xle9
$$

Divide entire inequality by $(3)$ as follows:

$$
xle3 therefore text{The solution set is} (-infty, 3]
$$

$$
text{color{Brown}(a) $(12.5, infty)$ (b) $left(-infty, -dfrac{23}{8}right]$
\ \
(c) $(-infty, 2)$ (d) $(-infty, 3]$}
$$

Solve the following inequalities and set your solution in notation.

$$
color{#4257b2}text{(a)} -3<2x+1<9
$$

Add entire inequality by $(-1)$ as follows:

$$
-4<2x<8
$$

Divide entire inequality by $(2)$ as follows:

$$
-2<x<4 therefore text{The solution set is} (-2, 4)
$$

$$
color{#4257b2}text{(b)} 8le-x+8le9
$$

Add entire inequality by $(-8)$ as follows:

$$
0le-xle1
$$

Divide entire inequality by $(-1)$ as follows:

$color{#4257b2}diamond Note that: Dividing by a negative number reverses the inequality sign.$

$$
0ge xge1 therefore text{The solution set is} [0, 1]
$$

$$
color{#4257b2}text{(c)} 6+2xge0ge-10+2x
$$

Add entire inequality by $(-2x)$ as follows:

$$
6ge-2xge-10
$$

Divide entire inequality by $(-2)$ as follows:

$$
-3le xle5 therefore text{The solution set is} [-3, 5]
$$

$$
color{#4257b2}text{(d)} x+1<2x+7<x+5
$$

Add entire inequality by $(-x)$ as follows:

$$
1<x+7<5
$$

Add entire inequality by $(-7)$ as follows:

$$
-6<x<-2 therefore text{The solution set is} (-6, -2)
$$

$$
text{color{Brown}(a) $(-2, 4)$ (b) $[0, 1]$
\ \
(c) $[-3, 5]$ (d) $(-6, -2)$}
$$

A phone company offered two options, First one is unlimited call for $(34.95)$ dollar per month. Second one is fee $(20.95)$ dollar per month plus $(0.046)$ dollar for each call.

$$
text{color{#4257b2} When the unlimited option is better?}
$$

The difference between two fees is $34.95-20.95=14$ dollar

Since the number of minute for $(14)$ dollar is $=dfrac{14}{0.04}=350$ call.

So the unlimited option is better when the call more than $(350)$ call per month.

$$
text{color{Brown} The unlimited option is better when the call more than $(350)$ call per month.}
$$

Determine which interval is true for the following inequalities.

$$
color{#4257b2}text{(a)} (x+1)(x-2)(x+3)^2<0
$$

$$
x=-3, x=-1, x=2
$$

Interval of $(x)$ are:

$$
x<-3, -3<x<-1, -1<x2
$$

Determine the value of polynomial according each interval as follows:

For $x<-3 f(-4)=18$ For $-3<x<-1 f(-2)=4$

For $-1<x2 f(3)=144$

According the inequality $f(x)<0$, the true interval which make the inequality is true is:

$$
-1<x<2
$$

$$
color{#4257b2}text{(b)} dfrac{(x-4)(2x+3)}{5}gedfrac{2x+3}{5}
$$

Multiply entire inequality by $left(dfrac{5}{2x+3}right)$ as follows:

$$
x-4ge0 xge4
$$

Interval of $(x)$ are:

$$
xle4, xge4
$$

Determine the value of polynomial according each interval as follows:

For $xle4 f(3)=-1$ For $xge4 f(5)=1$

According the inequality $f(x)ge0$, the true interval which make the inequality is true is:

$$
xge4
$$

$$
color{#4257b2}text{(c)} -2(x-1)(2x+5)(x-7)>0
$$

$$
x=-dfrac{5}{2}, x=1, x=27
$$

Interval of $(x)$ are:

$$
x<-dfrac{5}{2}, -dfrac{5}{2}<x<1, 1<x7
$$

Determine the value of polynomial according each interval as follows:

For $x<-dfrac{5}{2} f(-3)=80$ For $-dfrac{5}{2}<x<1 f(0)=-70$

For $1<x7 f(8)=294$

According the inequality $f(x)<0$, the true interval which make the inequality is true are:

$$
x<-dfrac{5}{2} 1<x<7
$$

$$
color{#4257b2}text{(d)} x^3+x^2-21x+21le3x^2-2x+1
$$

Isolate the variables on the left side as follows:

$$
x^3+x^2-21x+21-3x^2+2x-1le0
$$

$$
x^3-2x^2-19x+20le0
$$

Find the value of $f(1)$ to know if it factored or not as follows:

$$
f(1)=1-2-19+20=0
$$

The $(x-1)$ is factored, so use theorem property to factor the polynomials as follows:

$$
1 | 1 -2 -19 20
$$

$$
1 -1 -20
$$

$$
1 -1 -20 0
$$

$$
x^3-2x^2-19x+20=(x-1)(x^2-x-20)
$$

$$
(x-1)(x-5)(x+4)
$$

$$
x=-4, x=1, x=5
$$

Interval of $(x)$ are:

$$
xle-4, -4le xle1, 1le xle5, xge5
$$

Determine the value of polynomial according each interval as follows:

For $xle-4 f(-5)=-60$ For $-4le xle1 f(0)=20$

For $1le xle5 f(2)=-18$ For $xge5 f(6)=50$

According the inequality $f(x)<0$, the true interval which make the inequality is true is:

$$
xle-4 1le xle5
$$

$$
text{color{Brown}(a) $-1<x<2$ (b) $xge4$
\ \
(c) $x<dfrac{-5}{2}, 1<x<7$
\ \
(d) $xle-4, 1le xle5$}
$$

Determine where the interval of the following function is positive and negative.

$$
color{#4257b2}f(x)=2x^4-2x^3-32x^2-40x
$$

Let know if he function is factored or not by substitution values of $(x)$ until to get zero as follows:

$$
f(1)=2-2-32-40 f(1)=-72
$$

$$
f(-1)=2+2-32+40 f(-1)=12
$$

$$
f(-2)=32+16-128+80 f(-2)=0
$$

The function is factored of $(x+2)$, so use the theorem property to get the others zeros as follows:

$$
-2 |2 -2 -32 -40
$$

$$
-4 12 40
$$

$$
2 -6 -20 0
$$

$$
2x^4-2x^3-32x^2-40x=(x+2)(2x^3-6x^2-20x)
$$

$$
2x^4-2x^3-32x^2-40x=2x(x+2)(x-5)(x+2)
$$

$$
x=-2, x=0, x=5
$$

Determine the value of polynomial according each interval as follows:

For $x<-2 f(-3)=48$ For $-2<x<0 f(-1)=12$

For $0<x5 f(6)=768$

Function has a positive value at intervals of $x<-2, -2<x5$

Function has a negative value at interval of $0<x<5$

$$
text{color{Brown}Function has a positive value at intervals of $x<-2, -2<x5$
\ \
Function has a negative value at interval of $0<x<5$}
$$

Using a graphing calculator we can sketch both functions and look for the region on the graph where the graph of the function $x^3-2x^2+x-3$ is above the function $2x^3+x^2-x+1$.

Hence, entering both polynomial functions $y_1=x^3-2x^2+x-3$ and $y_2=2x^3+x^2-x+1$ and pressing GRAPH we obtain two graphs as depicted in *Figure 1*.

*Figure 1*. The graphs of the functions

As we can see in this figure, up until the point $(-3.803,-90.706)$ the red curve depicting the function $x^3-2x^2+x-3$ is above the blue one depicting the function $2x^3+x^2-x+1$. Hence,
$$x^3-2x^2+x-3ge2x^3+x^2-x+1$$
over this region where $xle-3.806$.

$xle-3.806$

Estimate the intervals in [year and months] for the following expression which represent the volume of wood from $1993$ to $1997$ when the volume less than $185000$ cubic meter

$$
color{#4257b2}f(x)=1135x^4-8197x^3+15868x^2-2157x+176608 text{ Where},0le xle4
$$

Interval of $(x)$ are:

$$
0le xle4
$$

Determine the value of polynomial according each interval as follows:

For $f(1.5)=187156.562$

For $0le xle4 f(1)=183257$

The interval which has a volume less than $185000$ cubic meter is $0le xle1.5$

$$
text{color{Brown}$0le xle1.5$}
$$

Estimate the average rate of change at the interval at $(x=2, x=7)$ and instantaneous rate of change at $(x=5)$ for the following expression.

$$
color{#4257b2}text{(a)} f(x)=x^2-2x+3
$$

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

$$
f(7)=(7)^2-2(7)+3 f(7)=38
$$

Average rate of change $=dfrac{38-3}{7-2}=dfrac{35}{5}$

Average rate of change $=7$

$text{color{#4257b2}Note that}$: the instantaneous rate of change is calculated at ${f}'(x)$, so we used the ${f}'(x)$ as follows:

$$
{f}'(x)=2x-2
$$

Instantaneous rate of change at $(x=5)$ is equal

$$
{f}'(5)=(2cdot5)-2 {f}'(5)=8
$$

$$
color{#4257b2}text{(b)} h(x)=(x-3)(2x+1)
$$

$$
h(x)=2x^2-5x-3
$$

$$
f(2)=2(2)^2-5(2)-3 f(2)=-5
$$

$$
f(7)=2(7)^2-5(7)-3 f(7)=60
$$

Average rate of change $=dfrac{60+5}{7-2}=dfrac{65}{5}$

Average rate of change $=13$

$text{color{#4257b2}Note that}$: the instantaneous rate of change is calculated at ${h}'(x)$, so we used the ${h}'(x)$ as follows:

$$
{h}'(x)=4x-5
$$

Instantaneous rate of change at $(x=5)$ is equal

$$
{f}'(5)=(4cdot5)-5 {f}'(5)=15
$$

$$
color{#4257b2}text{(c)} g(x)=2x^3-5x
$$

$$
f(2)=2(2)^3-5(2) f(2)=6
$$

$$
f(7)=2(7)^3-5(7) f(7)=651
$$

Average rate of change $=dfrac{651-6}{7-2}=dfrac{645}{5}$

Average rate of change $=129$

$text{color{#4257b2}Note that}$: the instantaneous rate of change is calculated at ${g}'(x)$, so we used the ${g}'(x)$ as follows:

$$
{g}'(x)=6x^2-5
$$

Instantaneous rate of change at $(x=5)$ is equal

$$
{f}'(5)=6(5)^2-5 {f}'(5)=145
$$

$$
color{#4257b2}text{(d)} v(x)=-x^4+2x^2-5x+1
$$

$$
f(2)=-(2)^4+2(2)^2-5(2)+1 f(2)=-17
$$

$$
f(7)=-(7)^4+2(7)^2-5(7)+1 f(7)=-2347
$$

Average rate of change $=dfrac{-2347+17}{7-2}=-dfrac{2330}{5}$

Average rate of change $=-466$

$text{color{#4257b2}Note that}$: the instantaneous rate of change is calculated at ${v}'(x)$, so we used the ${v}'(x)$ as follows:

$$
{v}'(x)=-4x^3+4x-5
$$

Instantaneous rate of change at $(x=5)$ is equal

$$
{f}'(5)=-4(5)^3+4(5)-5 {f}'(5)=-485
$$

$$
text{color{Brown}(a) Average rate is $=7$ Instantaneous rate $=8$
\ \
(b) Average rate is $=13$ Instantaneous rate $=15$
\ \
(c) Average rate is $=129$ Instantaneous rate $=145$
\ \
(d) Average rate is $=-466$ Instantaneous rate $=-485$}
$$

$$
text{color{#4257b2}Consider the graph on the textbook to estimate the interval of instantaneous rate of change is positive, negative and zero}
$$

From the graph on the textbook,

The instantaneous rate of change is positive at the interval of $0<x<1.65$

The instantaneous rate of change is negative at the interval of $-1.65<x<0$

The instantaneous rate of change is equal zero at the interval of $x=0$

$$
text{color{Brown}The instantaneous rate of change is positive at the interval of $0<x<1.65$
\ \
The instantaneous rate of change is negative at the interval of $-1.65<x<0$
\ \
The instantaneous rate of change is equal zero at the interval of $x=0$}
$$

The projectile is modelled by the following expression $h(t)=-5t^2+25$. Answer for the following questions.

$$
text{color{#4257b2}(a) Find the point when the object hits the ground}
$$

Since the object hits the ground, so the height is equal zero, $h(t)=0$

$$
0=-5t^2+25 5t^2=25
$$

Divide both of sides by $(5)$ as follows:

$$
t^2=dfrac{25}{5} t^2=5
$$

Use square root property as follows:

$$
sqrt{t^2}=sqrt{5} t=pmsqrt{5}
$$

Is not reality the time be a negative value, so the correct time is $(t=2.236)$

$$
text{color{#4257b2}(b) Find the average rate of change from the point $(t=0)$ to the hits point to the ground}
$$

$$
h(t)=-5t^2+25 t=0, t=sqrt{5}
$$

$$
f(0)=-5(0)^2+25 f(0)=25
$$

$$
f(sqrt{5})=-5(sqrt{5})^2+25 f(sqrt{5})=0
$$

Average rate of change is $=dfrac{0-25}{sqrt{5}-0}=-dfrac{25}{sqrt{5}}$

$$
text{color{#4257b2}(c) Find the object speed at the point of impact }
$$

The object speed is decreased when impact to the ground as $left(dfrac{25}{sqrt{5}}right)$ m/s

$$
text{color{Brown}(a) $t=sqrt{5}$ (b) Average rate of change is $=-dfrac{25}{sqrt{5}}$
\ \
(c) The object speed is decreased at impact moment as $left(dfrac{25}{sqrt{5}}right)$}
$$

Consider the following expression $f(x)=2x^3+3x-1$. Answer for the following questions.

$$
text{color{#4257b2}(a) Find the average rate from point $(x=3)$ to point $(x=3.0001)$}
$$

$$
f(3)=2(3)^3+3(3)-1 f(3)=62
$$

$$
f(3.0001)=2(3.0001)^3+3(3.0001)-1 f(3.0001)=62.0057
$$

Average rate of change is $=dfrac{62.0057-62}{3.0001-3}=dfrac{0.0057}{0.0001}=57$

$$
text{color{#4257b2}(b) Find the average rate from point $(x=2.9999)$ to point $(x=3)$}
$$

$$
f(2.9999)=2(2.9999)^3+3(2.9999)-1 f(2.9999)=61.9943
$$

$$
f(3)=2(3)^3+3(3)-1 f(3)=62
$$

Average rate of change is $=dfrac{62-61.9943}{3-2.9999}=dfrac{0.0057}{0.0001}=57$

$$
text{color{#4257b2}(c) Why the answer is similar? Find the instantaneous rate at value of $x=3$}
$$

Because the difference between the points is very very small tat may be neglected.

The instantaneous rate of change at $(x=3)$

$text{color{#4257b2}Note that}$: the instantaneous rate of change is calculated at ${f}'(x)$, so we used the ${f}'(x)$ as follows:

$$
{f}'(x)=6x+3
$$

Instantaneous rate of change at $(x=3)$ is equal

$$
{f}'(3)=(6cdot3)+3 {t}'(3)=21
$$

$$
text{color{Brown}(a) $57$ (b) $57$ (c) $21$}
$$