## Calculus: Early Transcendentals (2nd Edition)

$\dfrac{f(x+h)-f(x)}{h}=3x^{2}+3xh+h^{2}$ $\dfrac{f(x)-f(a)}{x-a}=x^{2}+ax+a^{2}$
$f(x)=x^{3}+2$ $\textbf{Evaluate}$ $\dfrac{f(x+h)-f(x)}{h}$ First, substitute $x$ by $x+h$ in the given function and simplify to find $f(x+h)$: $f(x+h)=(x+h)^{3}+2=x^{3}+3x^{2}h+3xh^{2}+h^{3}+2$ Substitute $f(x+h)$ and $f(x)$ into the difference quotient formula and simplify: $\dfrac{f(x+h)-f(x)}{h}=\dfrac{x^{3}+3x^{2}h+3xh^{2}+h^{3}+2-(x^{3}+2)}{h}=...$ $...=\dfrac{x^{3}+3x^{2}h+3xh^{2}+h^{3}+2-x^{3}-2}{h}=...$ $...=\dfrac{3x^{2}h+3xh^{2}+h^{3}}{h}=...$ Take out common factor $h$ from the numerator and simplify again: $...=\dfrac{h(3x^{2}+3xh+h^{2})}{h}=3x^{2}+3xh+h^{2}$ $\textbf{Evaluate}$ $\dfrac{f(x)-f(a)}{x-a}$ Substitute $x$ by $a$ in $f(x)$ to find $f(a)$: $f(a)=a^{3}+2$ Substitute $f(a)$ and $f(x)$ into the difference quotient formula and simplify: $\dfrac{f(x)-f(a)}{x-a}=\dfrac{x^{3}+2-(a^{3}+2)}{x-a}=\dfrac{x^{3}+2-a^{3}-2}{x-a}=...$ $...=\dfrac{x^{3}-a^{3}}{x-a}=...$ Factor the numerator and simplify again: $...=\dfrac{(x-a)(x^{2}+ax+a^{2})}{x-a}=x^{2}+ax+a^{2}$