Problem 85 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

The simplified difference quotient is \(-4x - 2h -1 .\)

Step 1: Finding f(x+h)

First, replace every \(x\) in the function \(f(x)\) with \(x+h\), to get \(f(x+h) = -2(x+h)^{2} - (x+h) +3.\)

Step 2: Expanding f(x+h)

Next, expand the equation to get \(f(x+h) = -2(x^{2} + 2xh + h^{2}) - x - h +3. Simplify the equation further to get \(f(x+h) = -2x^{2} -4hx -2h^{2} -x -h+3.\)

Step 3: Setup the difference quotient

Now, return to the difference quotient, and replace \(f(x+h)\) and \(f(x)\) with their respective values, to get \(\frac{-2x^{2} -4hx -2h^{2} -x-h+3 - (-2x^{2} -x+3)}{h}\). This simplifies to \(\frac{-4hx -2h^{2} -h}{h}\).

Step 4: Simplify the difference quotient

Next, factor out \(h\) from the numerator to eliminate the denominator, the simplified difference quotient is \(-4x - 2h - 1 .\)