Problem 81 - Precalculus Essentials

Question

Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=-x^{2}+2 x+4$$

Step-by-Step Solution

Answer

The simplified difference quotient of the function \(f(x)=-x^{2}+2x+4\) is \(-2x-h+2\).

Step 1: Substitute (x+h) into the function

First, replace all instances of \(x\) with \((x+h)\) in the function \(f(x)=-x^{2}+2x+4\). This results in: \(f(x+h)=-((x+h)^{2})+2(x+h)+4\).

Step 2: Simplify the function f(x+h)

Expand and simplify the result from step 1. Using the formula \((a+b)^2=a^2+2ab+b^2\), the function \(f(x+h)\) simplifies to: \(-x^{2}-2hx-h^{2}+2x+2h+4\).

Step 3: Replace f(x+h) and f(x) in the difference quotient

Now, put the expression for \(f(x+h)\) and the given expression for \(f(x)\) back into the difference quotient formula \(\frac{f(x+h)-f(x)}{h}\), which then becomes: \(\frac{-x^{2}-2hx-h^{2}+2x+2h+4-(-x^{2}+2x+4)}{h}\).

Step 4: Simplify the fraction

Simplifying the numerator by removing like terms results in: \(-2xh-h^{2}+2h\). This simplifies further to: \((-2x-h+2)h\). Then divide each term by \(h\), simplifying it to: \(-2x-h+2\).