## Calculus: Early Transcendentals (2nd Edition)

$\dfrac{f(x+h)-f(x)}{h}=-\dfrac{7}{(x+3)(x+h+3)}$ $\dfrac{f(x)-f(a)}{x-a}=-\dfrac{7}{(x+3)(a+3)}$
$f(x)=\dfrac{7}{x+3}$ $\textbf{Evaluate}$ $\dfrac{f(x+h)-f(x)}{h}$ First, substitute $x$ by $x+h$ in the given function and simplify to find $f(x+h)$: $f(x+h)=\dfrac{7}{x+h+3}$ Substitute $f(x+h)$ and $f(x)$ into the difference quotient formula and simplify: $\dfrac{f(x+h)-f(x)}{h}=\dfrac{\dfrac{7}{x+h+3}-\dfrac{7}{x+3}}{h}=...$ $...=\dfrac{\dfrac{7(x+3)-7(x+h+3)}{(x+3)(x+h+3)}}{h}=...$ $...=\dfrac{7x+21-7x-7h-21}{h(x+3)(x+h+3)}=-\dfrac{7h}{h(x+3)(x+h+3)}=...$ $...=-\dfrac{7}{(x+3)(x+h+3)}$ $\textbf{Evaluate}$ $\dfrac{f(x)-f(a)}{x-a}$ Substitute $x$ by $a$ in $f(x)$ to find $f(a)$: $f(a)=\dfrac{7}{a+3}$ Substitute $f(a)$ and $f(x)$ into the difference quotient formula and simplify: $\dfrac{f(x)-f(a)}{x-a}=\dfrac{\dfrac{7}{x+3}-\dfrac{7}{a+3}}{x-a}=\dfrac{\dfrac{7(a+3)-7(x+3)}{(x+3)(a+3)}}{x-a}=...$ $...=\dfrac{7a+21-7x-21}{(x-a)(x+3)(a+3)}=\dfrac{7a-7x}{(x-a)(x+3)(a+3)}=...$ Take out common factor $-7$ from the numerator and simplify again: $...=-\dfrac{7(x-a)}{(x-a)(x+3)(a+3)}=-\dfrac{7}{(x+3)(a+3)}$