chapter review exercises
1. the position of a particle at time $t(s)$ is $s(t)=sqrt{t^{2}+1}$ m. compute its average velocity over $2,5$ and estimate its instantaneous velocity at $t = 2$.
2. a rock dropped from a state of rest at time $t = 0$ on the planet ginormon travels a distance $s(t)=15.2t^{2}$ m in $t$ seconds. estimate the instantaneous velocity at $t = 5$.
3. for $f(x)=sqrt{2x}$ compute the slopes of the secant lines from 16 to each of $16pm0.01,16pm0.001,16pm0.0001$ and use those values to estimate the slope of the tangent line at $x = 16$.
4. show that the slope of the secant line for $f(x)=x^{3}-2x$ over $5,x$ is equal to $x^{2}+5x + 23$. use this to estimate the slope of the tangent line at $x = 5$.
in exercises 5 - 10, estimate the limit numerically to two decimal places or state that the limit does not exist.
5. $lim_{x
ightarrow0}\frac{1-cos^{3}(x)}{x^{2}}$
6. $lim_{x
ightarrow1}x^{1/(x - 1)}$
7. $lim_{x
ightarrow2}\frac{x^{x}-4}{x^{2}-4}$
8. $lim_{x
ightarrow2}\frac{x - 2}{ln(3x - 5)}$
9. $limleft(\frac{7}{} -\frac{3}{}
ight)$

Question

Accepted Answer

### Question 1
# Explanation:
## Step1: Recall average - velocity formula
The average velocity of a particle with position function $s(t)$ over the interval $[a,b]$ is $v_{avg}=\frac{s(b)-s(a)}{b - a}$. Given $s(t)=\sqrt{t^{2}+1}$, $a = 2$, and $b = 5$.
First, find $s(2)$ and $s(5)$:
$s(2)=\sqrt{2^{2}+1}=\sqrt{4 + 1}=\sqrt{5}$
$s(5)=\sqrt{5^{2}+1}=\sqrt{25+1}=\sqrt{26}$
Then $v_{avg}=\frac{s(5)-s(2)}{5 - 2}=\frac{\sqrt{26}-\sqrt{5}}{3}\approx\frac{5.099 - 2.236}{3}=\frac{2.863}{3}\approx0.95$ m/s.
## Step2: Recall instantaneous - velocity formula
The instantaneous velocity $v(t)$ is the derivative of the position function $s(t)$. Using the chain - rule, if $s(t)=(t^{2}+1)^{\frac{1}{2}}$, then $s^\prime(t)=\frac{1}{2}(t^{2}+1)^{-\frac{1}{2}}\cdot2t=\frac{t}{\sqrt{t^{2}+1}}$.
To find the instantaneous velocity at $t = 2$, substitute $t = 2$ into $s^\prime(t)$:
$s^\prime(2)=\frac{2}{\sqrt{2^{2}+1}}=\frac{2}{\sqrt{5}}\approx\frac{2}{2.236}\approx0.89$ m/s.

### Question 2
# Explanation:
## Step1: Recall instantaneous - velocity formula
The instantaneous velocity $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=15.2t^{2}$, using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, where $a = 15.2$ and $n = 2$.
So $v(t)=s^\prime(t)=30.4t$.
## Step2: Evaluate at $t = 5$
Substitute $t = 5$ into $v(t)$:
$v(5)=30.4	imes5 = 152$ m/s.

### Question 3
# Explanation:
## Step1: Recall slope of secant - line formula
The slope of the secant line between two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ on the curve $y = f(x)$ is $m=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Given $f(x)=\sqrt{2x}$, $x_1 = 16$.
For $x_2=16 + 0.01$:
$f(16)=\sqrt{2	imes16}=\sqrt{32}=4\sqrt{2}$
$f(16 + 0.01)=\sqrt{2	imes(16 + 0.01)}=\sqrt{32.02}$
$m_1=\frac{\sqrt{32.02}-4\sqrt{2}}{(16 + 0.01)-16}=\frac{\sqrt{32.02}-4\sqrt{2}}{0.01}\approx\frac{5.6586 - 5.6569}{0.01}=\frac{0.0017}{0.01}=0.17$
Similarly, for $x_2=16-0.01$:
$m_2=\frac{\sqrt{31.98}-4\sqrt{2}}{(16 - 0.01)-16}=\frac{\sqrt{31.98}-4\sqrt{2}}{-0.01}\approx0.17$
For $x_2=16 + 0.001$:
$m_3=\frac{\sqrt{32.002}-4\sqrt{2}}{0.001}\approx0.177$
For $x_2=16-0.001$:
$m_4=\frac{\sqrt{31.998}-4\sqrt{2}}{-0.001}\approx0.177$
For $x_2=16 + 0.0001$:
$m_5=\frac{\sqrt{32.0002}-4\sqrt{2}}{0.0001}\approx0.177$
For $x_2=16-0.0001$:
$m_6=\frac{\sqrt{31.9998}-4\sqrt{2}}{-0.0001}\approx0.177$
The slope of the tangent line at $x = 16$ is approximately $0.18$.

### Question 4
# Explanation:
## Step1: Recall slope of secant - line formula
The slope of the secant line between two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ on the curve $y = f(x)$ is $m=\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. Given $f(x)=x^{3}-2x$, $x_1 = 5$, and $x_2=x$.
$f(5)=5^{3}-2	imes5=125 - 10 = 115$
$f(x)=x^{3}-2x$
$m=\frac{(x^{3}-2x)-115}{x - 5}=\frac{x^{3}-2x-115}{x - 5}$
By polynomial long - division or synthetic division, $\frac{x^{3}-2x-115}{x - 5}=x^{2}+5x + 23$.
## Step2: Recall slope of tangent - line formula
The slope of the tangent line at $x = 5$ is the limit of the slope of the secant line as $x$ approaches $5$. Substitute $x = 5$ into $x^{2}+5x + 23$:
$m=5^{2}+5	imes5+23=25 + 25+23=73$.

### Question 5
# Explanation:
## Step1: Use L'Hopital's rule (since it is in $\frac{0}{0}$ form)
As $x
ightarrow0$, $\frac{1-\cos^{3}(x)}{x^{2}}$ is in the $\frac{0}{0}$ form (because $\cos(0)=1$, so $1-\cos^{3}(0)=0$ and $x^{2}=0$ when $x = 0$).
Differentiate the numerator and denominator. The derivative of the numerator $u = 1-\cos^{3}(x)$ using the chain - rule: $u^\prime=3\cos^{2}(x)\sin(x)$.
The derivative of the denominator $v=x^{2}$ is $v^\prime = 2x$.
So $\lim_{x
ightarrow0}\frac{1-\cos^{3}(x)}{x^{2}}=\lim_{x
ightarrow0}\frac{3\cos^{2}(x)\sin(x)}{2x}$.
This is still in the $\frac{0}{0}$ form. Apply L'Hopital's rule again.
The derivative of the numerator $3\cos^{2}(x)\sin(x)$ using the product - rule:
$(3\cos^{2}(x)\sin(x))^\prime=3(-2\cos(x)\sin^{2}(x)+\cos^{3}(x))$.
The derivative of the denominator $2x$ is $2$.
$\lim_{x
ightarrow0}\frac{3(-2\cos(x)\sin^{2}(x)+\cos^{3}(x))}{2}=\frac{3(0 + 1)}{2}=1.50$.

### Question 6
# Explanation:
## Step1: Let $y=x^{\frac{1}{x - 1}}$, then $\ln y=\frac{\ln x}{x - 1}$
As $x
ightarrow1$, $\frac{\ln x}{x - 1}$ is in the $\frac{0}{0}$ form (since $\ln(1)=0$ and $x - 1=0$ when $x = 1$).
Apply L'Hopital's rule. The derivative of the numerator $\ln x$ is $\frac{1}{x}$, and the derivative of the denominator $x - 1$ is $1$.
So $\lim_{x
ightarrow1}\frac{\ln x}{x - 1}=\lim_{x
ightarrow1}\frac{1}{x}=1$.
Since $\lim_{x
ightarrow1}\ln y = 1$, then $\lim_{x
ightarrow1}y=e^{1}\approx2.72$.

### Question 7
# Explanation:
## Step1: Factor the denominator
As $x
ightarrow2$, $\frac{x^{x}-4}{x^{2}-4}$ is in the $\frac{0}{0}$ form (since $2^{2}=4$ and $x^{2}-4=(x - 2)(x + 2)$).
We know that $x^{x}=e^{x\ln x}$. Using L'Hopital's rule, the derivative of the numerator $u = x^{x}=e^{x\ln x}$, and by the chain - rule and product - rule, $u^\prime=x^{x}(\ln x + 1)$.
The derivative of the denominator $v=x^{2}-4$ is $v^\prime = 2x$.
$\lim_{x
ightarrow2}\frac{x^{x}-4}{x^{2}-4}=\lim_{x
ightarrow2}\frac{x^{x}(\ln x + 1)}{2x}=\frac{2^{2}(\ln 2+1)}{2	imes2}=\frac{4(\ln 2 + 1)}{4}=\ln 2+1\approx0.69+1 = 1.69$.

### Question 8
# Explanation:
## Step1: Use L'Hopital's rule (since it is in $\frac{0}{0}$ form)
As $x
ightarrow2$, $\frac{x - 2}{\ln(3x - 5)}$ is in the $\frac{0}{0}$ form (because when $x = 2$, $x - 2=0$ and $\ln(3	imes2-5)=\ln(1)=0$).
Differentiate the numerator and denominator. The derivative of the numerator $u=x - 2$ is $u^\prime = 1$.
The derivative of the denominator $v=\ln(3x - 5)$ using the chain - rule is $v^\prime=\frac{3}{3x - 5}$.
$\lim_{x
ightarrow2}\frac{x - 2}{\ln(3x - 5)}=\lim_{x
ightarrow2}\frac{1}{\frac{3}{3x - 5}}=\lim_{x
ightarrow2}\frac{3x - 5}{3}=\frac{3	imes2-5}{3}=\frac{1}{3}\approx0.33$.

### Question 9
# Explanation:
## Step1: First, find a common denominator
$\frac{7}{x - 7}-\frac{3}{x - 3}=\frac{7(x - 3)-3(x - 7)}{(x - 7)(x - 3)}=\frac{7x-21-3x + 21}{(x - 7)(x - 3)}=\frac{4x}{(x - 7)(x - 3)}$.
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{4x}{(x - 7)(x - 3)}=\lim_{x
ightarrow\infty}\frac{4x}{x^{2}-10x + 21}$.
Divide both the numerator and denominator by $x^{2}$:
$\lim_{x
ightarrow\infty}\frac{\frac{4}{x}}{1-\frac{10}{x}+\frac{21}{x^{2}}}=0$.

# Answer:
1. Average velocity: $\approx0.95$ m/s, Instantaneous velocity at $t = 2$: $\approx0.89$ m/s
2. Instantaneous velocity at $t = 5$: $152$ m/s
3. Slope of tangent line at $x = 16$: $\approx0.18$
4. Slope of tangent line at $x = 5$: $73$
5. $\lim_{x
ightarrow0}\frac{1-\cos^{3}(x)}{x^{2}}=1.50$
6. $\lim_{x
ightarrow1}x^{\frac{1}{x - 1}}\approx2.72$
7. $\lim_{x
ightarrow2}\frac{x^{x}-4}{x^{2}-4}\approx1.69$
8. $\lim_{x
ightarrow2}\frac{x - 2}{\ln(3x - 5)}\approx0.33$
9. $\lim_{x
ightarrow\infty}(\frac{7}{x - 7}-\frac{3}{x - 3})=0$

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Recall average - velocity formula

Step2: Recall instantaneous - velocity formula

Question 2

Step1: Recall instantaneous - velocity formula

Step2: Evaluate at $t = 5$

Question 3

Step1: Recall slope of secant - line formula

Question 4

Answer: