<s>[INST] Are the extrema of this function global or local? Last question about this function, I promise. 
The function $f: \mathbb R \rightarrow \mathbb R$ is given by $$f(x) = \begin{cases} 
\frac{x^2+5x+7}{x+3} & \mathrm{for} \; x < -3 \\
0 & \mathrm{for} \; x = -3 \\
\frac{x^2+5x+7}{x+3} & \mathrm{for} \; -3 < x < -2 \\
1 & \mathrm{for} \; x = -2 \\
-x-e^{-x}+e^2-1 & \mathrm{for} \; x > -2
 \end{cases}$$
Find all extrema of the function, and also determine, without the use of a calculator, whether they are global or local. 

My Attempt:
The derivative is given by:
$$f'(x) = \begin{cases} 
\frac{x^2+6x+8}{(x+3)^2} & \mathrm{for} \; x < -3 \\
\mathrm{undefined} & \mathrm{for} \; x = -3 \\
\frac{x^2+6x+8}{(x+3)^2} & \mathrm{for} \; -3 < x < -2 \\
\mathrm{undefined} & \mathrm{for} \; x = -2 \\
-1+e^{-x} & \mathrm{for} \; x > -2
 \end{cases}$$
Solve the first one:
\begin{align*}
\frac{x^2+6x+8}{x^2+6x+9} &= 0 \\
x^2+6x+8 &= 0 \\
(x+2)(x+4) &= 0 \\
x=-2 &\vee x=-4
\end{align*}
Only $x=-4$ is in the domain. The second derivative is $$f''(x)=\frac{2x+6}{(x+3)^4}$$ Substitute in $x=-4$ to get $f''(x)=-2$, so it is a minimum. 
Solve the second one: \begin{align*} -1+e^{-x} &= 0 \\ e^{-x} &= 1 \\ -x&=0 \\ x &= 0
\end{align*}
Which is in the domain. Taking the second derivative gives $-e^{-x}$, and $-e^{-0}=-1$, so it is a minimum. 

How to determine whether the minima are global or local?
Thank you in advance. 

Edit: They are maxima, not minima. 
 [/INST] Hint: to solve the first one multiply both sides by $(x+3)^2$ to get a quadratic equation.
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