Problem 85 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

The difference quotient of the function \(f(x)=x^{2}-x+1\) for \(x=2\) and \(h \neq 0\), when simplified, is \(h+4+\frac{2}{h}\). For \(h\) approaching zero, the difference quotient evaluates to 4.

Step 1: Substituting \(2+h\) into the function

First, substitute \(x=2+h\) into the function \(f(x)=x^{2}-x+1\), which results in: \(f(2+h)=(2+h)^{2}-(2+h)+1\), and simplifying gives \(f(2+h)=h^{2}+4h+3\).

Step 2: Substituting \(2\) into the function

Next, substitute \(x=2\) into the function \(f(x)=x^{2}-x+1\), which results in: \(f(2)=(2)^{2}-(2)+1\), and simplifying gives \(f(2)=1\).

Step 3: Applying the Difference Quotient Formula

Substitute \(f(2+h)\) and \(f(2)\) into the difference quotient formula \(\frac{f(2+h)-f(2)}{h}\), and then simplify: \(\frac{(h^{2}+4h+3)-1}{h}=h+4+\frac{2}{h}\). Note: As \(h\) approaches zero, \(\frac{2}{h}\) will also approach zero.