## Calculus: Early Transcendentals (2nd Edition)

$\dfrac{f(x+h)-f(x)}{h}=2x+h-2$ $\dfrac{f(x)-f(a)}{x-a}=x+a-2$
$f(x)=x^{2}-2x$ $\textbf{Evaluate}$ $\dfrac{f(x+h)-f(x)}{h}$ First, substitute $x$ by $x+h$ in the given function and simplify to find $f(x+h)$: $f(x+h)=(x+h)^{2}-2(x+h)=x^{2}+2xh+h^{2}-2x-2h$ Substitute $f(x+h)$ and $f(x)$ into the difference quotient formula and simplify: $\dfrac{f(x+h)-f(x)}{h}=\dfrac{x^{2}+2xh+h^{2}-2x-2h-(x^{2}-2x)}{h}=...$ $...=\dfrac{x^{2}+2xh+h^{2}-2x-2h-x^{2}+2x}{h}=...$ $...=\dfrac{2xh+h^{2}-2h}{h}=...$ Take out common factor $h$ from the numerator and simplify again: $...=\dfrac{h(2x+h-2)}{h}=2x+h-2$ $\textbf{Evaluate}$ $\dfrac{f(x)-f(a)}{x-a}$ Substitute $x$ by $a$ in $f(x)$ to find $f(a)$: $f(a)=a^{2}-2a$ Substitute $f(a)$ and $f(x)$ into the difference quotient formula and simplify: $\dfrac{f(x)-f(a)}{x-a}=\dfrac{x^{2}-2x-(a^{2}-2a)}{x-a}=\dfrac{x^{2}-2x-a^{2}+2a}{x-a}=...$ Factor the numerator by grouping. The first group is a difference of squares and the second group can be factored taking out common factor $2$: $...=\dfrac{(x^{2}-a^{2})-(2x-2a)}{x-a}=\dfrac{(x-a)(x+a)-2(x-a)}{x-a}=...$ Take out common factor $x-a$ from the numerator and simplify again: $...=\dfrac{(x-a)(x+a-2)}{x-a}=x+a-2$