Problem 79 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

The simplified difference quotient for the function \(f(x) = 2x^{2} + x - 1\) is \(4x + 2h + 1\).

Step 1: Substitute the function into the difference quotient formula

We replace \(f(x)\) and \(f(x+h)\) in the difference quotient formula \[\frac{f(x+h)-f(x)}{h}\] with the given function \(f(x) = 2x^{2} + x - 1\). This yields: \[\frac{2(x+h)^{2} + (x+h) - 1 - (2x^{2} + x - 1)}{h}\]

Step 2: Simplify the numerator

In this step, the goal is to simplify the equation above by expanding and combining like terms in the numerator. Starting with the expression \(2(x+h)^{2} = 2(x^{2} + 2xh + h^{2}) = 2x^{2} + 4xh + 2h^{2}\) and substituting it in, we get:\[\frac{2x^{2} + 4xh + 2h^{2} + x + h - 1 - 2x^{2} - x + 1}{h}\]Simplifying further gives:\[\frac{4xh + 2h^{2} + h}{h}\]

Step 3: Simplify the quotient

Now that the numerator is simplified, we can divide each term in the numerator by \(h\) (since \(h \neq 0\)) to complete the simplification of the difference quotient:\[\frac{4xh + 2h^{2} + h}{h} = 4x + 2h + 1\]