Question

given $f(x)=\frac{1}{x + 8}$, find the average rate of change of $f(x)$ on the interval $3, 3 + h$. your answer will be an expression involving $h$.

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \( f(x) \) on the interval \([a, b]\) is given by \(\frac{f(b)-f(a)}{b - a}\). Here, \( a = 3 \) and \( b=3 + h \), so we need to find \( f(3 + h) \) and \( f(3) \) first.

Step2: Calculate \( f(3) \)

Substitute \( x = 3 \) into \( f(x)=\frac{1}{x + 8} \):
\( f(3)=\frac{1}{3 + 8}=\frac{1}{11} \)

Step3: Calculate \( f(3 + h) \)

Substitute \( x = 3 + h \) into \( f(x)=\frac{1}{x + 8} \):
\( f(3 + h)=\frac{1}{(3 + h)+8}=\frac{1}{11 + h} \)

Step4: Apply the average rate of change formula

The average rate of change is \(\frac{f(3 + h)-f(3)}{(3 + h)-3}\). Substitute the values of \( f(3 + h) \) and \( f(3) \):
\[

$$\begin{align*} \frac{f(3 + h)-f(3)}{(3 + h)-3}&=\frac{\frac{1}{11 + h}-\frac{1}{11}}{h}\\ &=\frac{\frac{11-(11 + h)}{11(11 + h)}}{h}\\ &=\frac{\frac{11 - 11 - h}{11(11 + h)}}{h}\\ &=\frac{\frac{-h}{11(11 + h)}}{h}\\ &=\frac{-h}{11(11 + h)}\times\frac{1}{h}\\ &=\frac{-1}{11(11 + h)} \end{align*}$$

\]
(Note: \( h
eq0 \) to cancel the \( h \) terms)

Answer:

\(\frac{-1}{11(11 + h)}\) (or \(-\frac{1}{121 + 11h}\))