Question

question 6
given that $s(t)=12t - t^{2}$ is a position function where $s(t)$ is measured in millimeters and $t$ is time in seconds, find the average velocity (average rate of change) for $s(t)$ between $t = 1$ and $t = 3$ seconds.
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*algebraic work must be shown to retain points on quizzes, midterm or final
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question 7
calculate the average rate of change between $x = 9$ and $x = 8$ for the function $f(x)=1+\frac{9}{7}x$.
average rate of change =
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question 8
given the function $f(x)=8x^{2}-7x - 5$, find and simplify the difference quotient.

Explanation:

Question 6

Step1: Recall average - rate - of - change formula

The average rate of change of a function $s(t)$ over the interval $[a,b]$ is $\frac{s(b)-s(a)}{b - a}$. Here, $a = 1$, $b = 3$, and $s(t)=12t - t^{2}$.

Step2: Calculate $s(3)$

Substitute $t = 3$ into $s(t)$: $s(3)=12\times3-3^{2}=36 - 9=27$.

Step3: Calculate $s(1)$

Substitute $t = 1$ into $s(t)$: $s(1)=12\times1-1^{2}=12 - 1 = 11$.

Step4: Calculate the average rate of change

$\frac{s(3)-s(1)}{3 - 1}=\frac{27-11}{2}=\frac{16}{2}=8$ millimeters per second.

Step1: Recall average - rate - of change formula

The average rate of change of a function $f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 8$, $b = 9$, and $f(x)=1+\frac{9}{7}x$.

Step2: Calculate $f(9)$

$f(9)=1+\frac{9}{7}\times9=1+\frac{81}{7}=\frac{7 + 81}{7}=\frac{88}{7}$.

Step3: Calculate $f(8)$

$f(8)=1+\frac{9}{7}\times8=1+\frac{72}{7}=\frac{7+72}{7}=\frac{79}{7}$.

Step4: Calculate the average rate of change

$\frac{f(9)-f(8)}{9 - 8}=\frac{\frac{88}{7}-\frac{79}{7}}{1}=\frac{\frac{88 - 79}{7}}{1}=\frac{9}{7}$.

Step1: Recall the difference - quotient formula

The difference quotient of a function $f(x)$ is $\frac{f(x + h)-f(x)}{h}$, where $f(x)=8x^{2}-7x - 5$. First, find $f(x + h)$.

Step2: Calculate $f(x + h)$

$f(x + h)=8(x + h)^{2}-7(x + h)-5=8(x^{2}+2xh+h^{2})-7x-7h - 5=8x^{2}+16xh+8h^{2}-7x-7h - 5$.

Step3: Calculate $f(x + h)-f(x)$

\[

$$\begin{align*} f(x + h)-f(x)&=(8x^{2}+16xh+8h^{2}-7x-7h - 5)-(8x^{2}-7x - 5)\\ &=8x^{2}+16xh+8h^{2}-7x-7h - 5 - 8x^{2}+7x + 5\\ &=16xh+8h^{2}-7h \end{align*}$$

\]

Step4: Calculate the difference quotient

$\frac{f(x + h)-f(x)}{h}=\frac{16xh+8h^{2}-7h}{h}=\frac{h(16x + 8h-7)}{h}=16x+8h - 7$.

Answer:

$8$

Question 7