Problem

Find the difference quotient $\frac{f(x+h)-f(x)}{h}$, where $h \neq 0$, for the function below. \[ f(x)=\frac{1}{x-8} \] Simplify your answer as much as possible. \[ \frac{f(x+h)-f(x)}{h}=\square \]

Solution

Step 1 :The difference quotient is a measure of the average rate of change of the function over the interval h. To find the difference quotient for the function \(f(x)=\frac{1}{x-8}\), we need to substitute \(x+h\) into the function for \(x\), subtract the original function, and then divide by \(h\).

Step 2 :Substitute \(x+h\) into the function for \(x\): \(f(x+h) = \frac{1}{(x+h)-8}\)

Step 3 :Subtract the original function: \(f(x+h) - f(x) = \frac{1}{(x+h)-8} - \frac{1}{x-8}\)

Step 4 :Divide by \(h\): \(\frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{(x+h)-8} - \frac{1}{x-8}}{h}\)

Step 5 :Simplify the expression to get the difference quotient: \(\frac{f(x+h)-f(x)}{h} = -\frac{1}{(x - 8)(h + x - 8)}\)

Step 6 :Final Answer: \(\boxed{-\frac{1}{(x - 8)(h + x - 8)}}\)

From Solvely APP
Source: https://solvelyapp.com/problems/18523/

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