Find and simplify the difference quotient (f(x + h) - f(x))/h, where h is not zero, for f(x) = 2x^2 - 3x + 4.

Step-by-step solution

1. 1

   Build f(x + h) by substituting the whole expression x + h. Every $x$ in the rule is replaced, brackets included:

   $f(x+h)=2(x+h)^{2}-3(x+h)+4.$

2. 2

   Expand carefully. The square is the step where the middle term is often dropped: $(x+h)^{2}=x^{2}+2xh+h^{2}$, so

   $f(x+h)=2x^{2}+4xh+2h^{2}-3x-3h+4.$

3. 3

   Subtract f(x) and watch the cancellation. With $f(x)=2x^{2}-3x+4$, every term that does not contain $h$ disappears:

   $f(x+h)-f(x)=\left(2x^{2}-2x^{2}\right)+4xh+2h^{2}+\left(-3x+3x\right)-3h+\left(4-4\right)=4xh+2h^{2}-3h.$

   If anything without an $h$ survives here, there is an algebra error — the numerator must be divisible by $h$.

4. 4

   Factor h out of the numerator.

   $4xh+2h^{2}-3h=h\left(4x+2h-3\right).$

5. 5

   Cancel the h. This is legitimate precisely because $h\neq 0$ is given:

   $\frac{h\left(4x+2h-3\right)}{h}=4x+2h-3.$

6. 6

   Interpret the result. Letting $h\to 0$ gives $4x-3$, which is exactly $f'(x)$ — the difference quotient is the derivative before the limit is taken. That is a fast way to check the answer: differentiate $2x^2-3x+4$ term by term and confirm you get $4x-3$.