Problem 84 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

-6x - 3h + 2

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

First, replace \(x\) with \((x+h)\) in the original function \(f(x)\), obtaining \(f(x+h)= -3(x+h)^{2} + 2(x+h) - 1\). This equals to \(-3(x^{2}+2xh+h^{2}) + 2x + 2h - 1 = -3x^{2} - 6xh - 3h^{2} + 2x + 2h - 1 \) after expanding.

Step 2: Apply the Difference Quotient

Now, substitute \(f(x)\) and \(f(x+h)\) into the difference quotient formula, \(\frac{f(x+h) - f(x)}{h}\), which will look like this: \(\frac{-3x^{2} - 6xh - 3h^{2} + 2x + 2h - 1 - (-3x^{2} + 2x -1)}{h}\).

Step 3: Simplify the Difference Quotient

Next, simplify the above expression by cancelling out the \(3x^{2}\), \(2x\), and \(-1\) terms, and factor out an \(h\) from the remaining terms, which will look like this: \(\frac{-6xh - 3h^{2} + 2h}{h} = -6x - 3h + 2\). When \(h \neq 0\), this simplifies to \(-6x - 3h + 2\).