consider the graph of the function $f(x)=-\frac{1}{x + 3}+2$. compute $f(-2)$ $f(-2)$

Explanation:

Step1: Calculate $f(-2)$

Substitute $x = - 2$ into $f(x)=-\frac{1}{x + 3}+2$. [ \begin{align*} f(-2)&=-\frac{1}{-2 + 3}+2\ &=-\frac{1}{1}+2\ &=-1 + 2\ &=1 \end{align*} ]

Step2: Differentiate $f(x)$

First, rewrite $f(x)=-(x + 3)^{-1}+2$. Using the power - rule for differentiation $\frac{d}{dx}(u^n)=nu^{n - 1}\frac{du}{dx}$, where $u=x + 3$ and $n=-1$. The derivative of $-(x + 3)^{-1}$ is $(-1)\times(-1)\times(x + 3)^{-2}=\frac{1}{(x + 3)^2}$, and the derivative of the constant 2 is 0. So $f'(x)=\frac{1}{(x + 3)^2}$.

Step3: Calculate $f'(-2)$

Substitute $x=-2$ into $f'(x)$. [ \begin{align*} f'(-2)&=\frac{1}{(-2 + 3)^2}\ &=\frac{1}{1^2}\ &=1 \end{align*} ]

Answer:

$f(-2)=1$ $f'(-2)=1$