Question

question
find the slope of the tangent line to the graph of $f(x)=\frac{2}{x + 1}$ at $x = 2$.
provide your answer below:
$m_{tan}=square$

Explanation:

Step1: Recall slope - limit formula

The slope $m$ of the tangent line to the graph of $y = f(x)$ at $x=a$ is given by $m=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $f(x)=\frac{2}{x + 1}$ and $a = 2$.

Step2: Calculate $f(a+h)$ and $f(a)$

First, find $f(2+h)=\frac{2}{(2 + h)+1}=\frac{2}{h + 3}$, and $f(2)=\frac{2}{2 + 1}=\frac{2}{3}$.

Step3: Substitute into the limit formula

$m=\lim_{h
ightarrow0}\frac{\frac{2}{h + 3}-\frac{2}{3}}{h}=\lim_{h
ightarrow0}\frac{\frac{6-2(h + 3)}{3(h + 3)}}{h}=\lim_{h
ightarrow0}\frac{6-2h-6}{3h(h + 3)}$.

Step4: Simplify the expression

$\lim_{h
ightarrow0}\frac{-2h}{3h(h + 3)}=\lim_{h
ightarrow0}\frac{-2}{3(h + 3)}$.

Step5: Evaluate the limit

Substitute $h = 0$ into $\frac{-2}{3(h + 3)}$, we get $m=-\frac{2}{9}$.

Answer:

$-\frac{2}{9}$