91Ó°ÊÓ

The function:

$$\begin{gathered} f\left(x\right)=\sin \left(\frac{\mathrm{Ï€}}{2}x\right) \\ [-2,2] \end{gathered}$$

The critical points are given by,
$$\begin{gathered} f\left(x\right)=\sin \left(\frac{\mathrm{Ï€}}{2}x\right) \\ f\text{'}\left(x\right)=1 \end{gathered}$$

The critical points can tested as:

$f\text{'}\text{'}\left(x\right)=0$

Since the function does not have a solution, so the function does not have local maximum or local minimum.

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

$$\begin{gathered} f(-2)=\sin \left(\frac{\mathrm{Ï€}}{2}\right(-2\left)\right) \\ =0 \\ f\left(2\right)=\sin (\frac{\mathrm{Ï€}}{2}.2) \\ =0 \end{gathered}$$

The global maximum is at $x=1withf\left(1\right)=1$and the global minimum is at $x=-1withf(-1)=-1$.

The graph of the function is:

Find the global extrema in the interval $(-2,2)$.

$$\begin{gathered} \underset{x→-2^{-}}{lim}f\left(x\right)=\underset{x→2^{-}}{lim}\sin \left(\frac{\mathrm{π}}{2}x\right) \\ =0 \\ \underset{x→-2^{+}}{lim}f\left(x\right)=\underset{x→2^{+}}{lim}\sin \left(\frac{\mathrm{π}}{2}x\right) \\ =0 \end{gathered}$$

There is no global maximum and the global minimum.

The graph of the function is:

Find the global extrema in the interval $[-1,1)$.

$$\begin{gathered} f(-1)=\sin \left(\frac{\mathrm{π}}{2}\right(-1\left)\right) \\ =-1 \\ \underset{x→1^{+}}{lim}f\left(1\right)=\underset{x→1^{+}}{lim}\sin \left(\frac{\mathrm{π}}{2}\right(1\left)\right) \\ =1 \end{gathered}$$

There is no global maximum and the global minimum is at $x=-1withf(-1)=-1$.

The graph of the function is:

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

$$\begin{gathered} f\left(0\right)=\sin \left(\frac{\mathrm{π}}{2}\right(0\left)\right) \\ =-1 \\ \underset{x→{∞}^{+}}{lim}f\left(1\right)=\underset{x→{∞}^{+}}{lim}\sin \left(\frac{\mathrm{π}}{2}\right(∞\left)\right) \\ =∞ \end{gathered}$$

There is no global maximum and the global minimum is $x=-1withf(-1)=-1$.

The graph of the function is: