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3.3 rates of change and behavior of graphs score: 5/22 answered: 5/22 q…

Question

3.3 rates of change and behavior of graphs
score: 5/22 answered: 5/22
question 6
find the average rate of change on the interval specified for real numbers h (where h ≠ 0). simplify your answer.
g(x)=\frac{1}{x + 4} on 5,5 + h
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Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = g(x)$ on the interval $[a,b]$ is $\frac{g(b)-g(a)}{b - a}$. Here, $a = 5$, $b=5 + h$, and $g(x)=\frac{1}{x + 4}$.

Step2: Calculate $g(5 + h)$ and $g(5)$

$g(5 + h)=\frac{1}{(5 + h)+4}=\frac{1}{h + 9}$, and $g(5)=\frac{1}{5 + 4}=\frac{1}{9}$.

Step3: Substitute into the formula

$\frac{g(5 + h)-g(5)}{(5 + h)-5}=\frac{\frac{1}{h + 9}-\frac{1}{9}}{h}$.

Step4: Find a common denominator for the numerator

The common denominator of $\frac{1}{h + 9}$ and $\frac{1}{9}$ is $9(h + 9)$. So $\frac{1}{h + 9}-\frac{1}{9}=\frac{9-(h + 9)}{9(h + 9)}=\frac{9 - h - 9}{9(h + 9)}=\frac{-h}{9(h + 9)}$.

Step5: Simplify the complex - fraction

$\frac{\frac{-h}{9(h + 9)}}{h}=\frac{-h}{9(h + 9)}\cdot\frac{1}{h}=-\frac{1}{9(h + 9)}$.

Answer:

$-\frac{1}{9(h + 9)}$