find the difference quotient $\frac{f(x + h)-f(x)}{h}$, $h\neq0$, for the given function $f$…

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$: $f(x + h)=-\frac{1}{(x + h)+2}=-\frac{1}{x + h+2}$

Step2: Calculate $f(x + h)-f(x)$

[ \begin{align*} f(x + h)-f(x)&=-\frac{1}{x + h+2}-\left(-\frac{1}{x + 2}\right)\ &=-\frac{1}{x + h+2}+\frac{1}{x + 2}\ &=\frac{-(x + 2)+(x + h+2)}{(x + 2)(x + h+2)}\ &=\frac{-x-2+x + h+2}{(x + 2)(x + h+2)}\ &=\frac{h}{(x + 2)(x + h+2)} \end{align*} ]

Step3: Find the difference - quotient

[ \begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{\frac{h}{(x + 2)(x + h+2)}}{h}\ &=\frac{h}{(x + 2)(x + h+2)}\times\frac{1}{h}\ &=\frac{1}{(x + 2)(x + h+2)} \end{align*} ]

Answer:

$\frac{1}{(x + 2)(x + h+2)}$