Find the locations and values of any global extrema of each function f in Exercises 11–20 on each of the four given intervals. Do all work by hand by considering local extrema and endpoint behavior. Afterwards, check your answers with graphs.

$f\left(x\right)=e^x(x-2)$on the interval

$$\begin{gathered} \left(a\right)[-2,2] \\ \left(b\right)(0,3) \\ \left(c\right)[0,\infty ) \\ \left(d\right)(-\infty ,0] \end{gathered}$$

The function:

$$\begin{gathered} f\left(x\right)=e^x(x-2) \\ [-2,2] \end{gathered}$$

The critical points are given by,
$$\begin{gathered} f\left(x\right)=e^x(x-2) \\ f\text{'}\left(x\right)=e^x(x-1) \\ f\text{'}\left(x\right)=0 \\ {e}^x(x-1)=0 \\ x=0,1 \end{gathered}$$

The critical points can tested as:

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=x{e}^x \\ f\text{'}\text{'}\left(1\right)=2.71828>0 \end{gathered}$$

So the function has a local minimum at $x=1$and there is no local maximum.

The height of the local extrema is,

$$\begin{gathered} f\left(1\right)=e^1(1-2) \\ =-2.72 \end{gathered}$$

Find the global extrema in the interval $[-2,2]$.

$$\begin{gathered} f(-2)=e^{-2}\left(\right(-2)-2) \\ =-0.54 \\ f\left(2\right)=e^2(2-2) \\ =0 \end{gathered}$$

There is no global maximum and the global minimum is at $x=0withf\left(1\right)=-2.72$

The graph of the function is:

Find the global extrema in the interval $(0,3)$.

$$\begin{gathered} \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}{e}^x(x-2) \\ =-2 \\ \underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3^{-}}{lim}{e}^x(x-2) \\ =20.08 \end{gathered}$$

There is no global maximum and the global minimum.

The graph of the function is:

Find the global extrema in the interval $[0,\infty )$.

$$\begin{gathered} f\left(0\right)=e^0(0-2) \\ =-2 \\ \underset{x\to \infty -}{lim}f\left(0\right)=\underset{x\to \infty -}{lim}{e}^{\infty }(\infty -2) \\ =\infty \end{gathered}$$

There is no global maximum and the global minimum is at $x=-2withf\left(0\right)=-2$.

The graph of the function is:

Find the global extrema in the interval $(-\infty ,0]$.

$$\begin{gathered} \underset{x\to -{\infty }^{+}}{lim}f\left(x\right)=\underset{x\to -{\infty }^{+}}{lim}{e}^x(x-2) \\ =-\infty \\ f\left(0\right)=e^0(0-2) \\ =-2 \end{gathered}$$

There is no global maximum and the global minimum is $x=0withf\left(0\right)=-2$.

The graph of the function is: