Question

given that $f(x)=\frac{1}{x}$, determine an expression in terms of $x$ and $h$ that represents the average rate of change of $f$ over any interval of length $h$. that is, over any interval $(x,x + h)$. simplify your answer as much as possible.

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question 7. points possible: 2

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $(x,x + h)$ is $\frac{f(x + h)-f(x)}{h}$.

Step2: Find $f(x + h)$

Given $f(x)=\frac{1}{x}$, then $f(x + h)=\frac{1}{x + h}$.

Step3: Substitute into the formula

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

Step4: Get a common denominator for the numerator

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

Step5: Simplify the complex - fraction

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

Answer:

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