Problem 83 - Precalculus Essentials

Question

Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=-2 x^{2}+5 x+7$$

Step-by-Step Solution

Answer

The simplified difference quotient for the function \( f(x) = -2 x^{2}+5 x+7 \) is \( -4x -2h +5 \).

Step 1: Substitute the function into the difference quotient

First, substitute \( f(x) = -2 x^{2}+5 x+7 \) into the difference quotient \(\frac{f(x+h)-f(x)}{h}\), to obtain \( \frac{[-2 (x+h)^{2}+5 (x+h)+7] - [-2 x^{2}+5 x+7]}{h} \). This expresses the difference quotient in terms of the given function and its shifted version.

Step 2: Expand and simplify the numerator

Next, expand the brackets in the numerator. This yields \( \frac{-2 (x^{2}+2 xh+h^{2})+5 (x+h)+7 +2 x^{2}-5 x-7}{h} \). Now, combine like terms to simplify the numerator further, to get \( \frac{-2x^{2}-4xh-2h^{2}+5x+5h+7+2x^{2}-5x-7}{h} \). After simplification, the numerator becomes \( -4xh-2h^2+5h \).

Step 3: Cancel common factors

In the final step, cancel the common factor of \( h \) from the numerator and denominator. This gives the simplified difference quotient, which is \( -4x -2h +5 \).