Problem 82 - Precalculus Essentials

Question

Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=-x^{2}-3 x+1$$

Step-by-Step Solution

Answer

The simplified difference quotient for the function \(f(x)=-x^{2}-3 x+1\) is \(-2x-h-3\).

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

First, replace every \(x\) in the function \(f(x)=-x^{2}-3 x+1\) by \(x+h\). Applying this substitution gives us \(f(x+h)=-(x+h)^{2}-3(x+h)+1\). This needs to be expanded. Doing so yields \(f(x+h)=-(x^2+2hx+h^2)-3x-3h+1\

Step 2: Substitute back into the Difference Quotient

Next, substitute \(f(x+h)\) and \(f(x)\) back into the difference quotient formula \(\frac{f(x+h)-f(x)}{h}\), which becomes \(\frac{-(x^2+2hx+h^2)-3x-3h+1-(-x^{2}-3 x+1)}{h}\).

Step 3: Simplify

Simplify the expression. \(x^2\), \(-3x\), and \(1\) cancel each other out leaving us with \(\frac{-2hx-h^2-3h}{h}\). This can be further simplified by factoring out \(h\) in the numerator, giving the result \(\frac{h(-2x-h-3)}{h}\). After that, \(h\) in both the numerator and the denominator cancel each other out, resulting in our final simplified difference quotient \(-2x-h-3\).