Question

homework assignment 3.3 rates of change and behavior of graphs score: 4/11 answered: 4/11 question 6 given (f(x)=\frac{1}{x + 12}), find the average rate of change of (f(x)) on the interval (10,10 + h). your answer will be an expression involving (h). question help: video written example message instructor submit question

Explanation:

Step1: Recall average rate - of - change formula

The average rate of change of a function $y = f(x)$ on the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 10$, $b=10 + h$, and $f(x)=\frac{1}{x + 12}$.

Step2: Calculate $f(10 + h)$ and $f(10)$

$f(10 + h)=\frac{1}{(10 + h)+12}=\frac{1}{h + 22}$, and $f(10)=\frac{1}{10+12}=\frac{1}{22}$.

Step3: Substitute into the formula

The average rate of change is $\frac{f(10 + h)-f(10)}{(10 + h)-10}=\frac{\frac{1}{h + 22}-\frac{1}{22}}{h}$.

Step4: Simplify the numerator

$\frac{1}{h + 22}-\frac{1}{22}=\frac{22-(h + 22)}{22(h + 22)}=\frac{22 - h - 22}{22(h + 22)}=\frac{-h}{22(h + 22)}$.

Step5: Divide by $h$

$\frac{\frac{-h}{22(h + 22)}}{h}=\frac{-h}{22(h + 22)}\times\frac{1}{h}=-\frac{1}{22(h + 22)}$.

Answer:

$-\frac{1}{22(h + 22)}$