Q. 18 - Calculus

Question

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.

$f\left(x\right)=\sin \left(\frac{\mathrm{\pi }}{2}x\right)$ on the interval

$$\begin{gathered} \left(a\right) [-2,2] \\ \left(b\right) (-2,2) \\ \left(c\right) [-1,1) \\ \left(d\right) [0,\infty ) \end{gathered}$$

Step-by-Step Solution

Answer

(a) The global maximum of the function $f\left(x\right)=\sin \left(\frac{\mathrm{\pi }}{2}x\right)$ is  and at the values and at the values  and the global minimum at  and at the values .

(b) There is no global maximum and the global minimum. 

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

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

Part (a) Step 1. Given Information.

The function: 

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

Part (a) Step 2. Find the critical points.

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

Part (a) Step 3. Test the critical points.

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.

Part (a) Step 4. Check the height at endpoint values.

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

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

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

Part (a) Step 5. Sketch the graph.

The graph of the function is: 

Part (b) Step 1. Check the height at endpoint values.

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

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

There is no global maximum and the global minimum.

Part (b) Step 2. Graph the function.

The graph of the function is: 

Part (c) Step 1. Check the height at endpoint values.

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

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

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

Part (c) Step 2. Graph the function.

The graph of the function is: 

Part (d) Step 1. Check the height at endpoint values.

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

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

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

Part (d) Step 2. Graph the function.

The graph of the function is: