Q. 17 - 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.

$x^{\frac{3}{2}}(3x-5)$ on the interval

$$\begin{gathered} \left(a\right) [0,4] \\ \left(b\right) [0,4) \\ \left(c\right) [0,1) \\ \left(d\right) (0,1) \end{gathered}$$

Step-by-Step Solution

Answer

(a) The global maximum of the function $f\left(x\right)=x^{\frac{3}{2}}(3x-5)$ is $x=4$ and at the values $f\left(4\right)=56$ and the global minimum at $x=1$ and at the values $f\left(1\right)=-2$.

(b) The global minimum at $x=1$ and at the values $f\left(1\right)=-2$.

(c) The global maximum of the function is $x=0$ and at the values $f\left(0\right)=0$ and there is no global minimum.

(d) There is no global maximum and global minimum.

Part (a) Step 1. Given Information.

The function: 

$$\begin{gathered} f\left(x\right) =x^{\frac{3}{2}}(3x-5) \\ [0,4] \end{gathered}$$

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

The critical points are given by,
$$\begin{gathered} f\left(x\right)=x^{\frac{3}{2}}(3x-5) \\ f\text{'}\left(x\right)=x^{\frac{1}{2}}\left(\frac{15x-15}{2}\right) \end{gathered}$$

So,

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ {x}^{\frac{1}{2}}\left(\frac{15x-15}{2}\right)=0 \\ \sqrt{x}(15x-15)=0 \\ 15x=15 \\ x=1 \end{gathered}$$

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

The critical points can tested as:

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{4x-15}{4\sqrt{x}} \\ f\text{'}\text{'}\left(1\right)=-2.75<0 \end{gathered}$$ 

So the function has a local maximum at $x=1$

The height of the local extrema is, 

$$\begin{gathered} f\left(1\right)=1^{\frac{3}{2}}\left(3\right(1)-5) \\ =-2 \end{gathered}$$

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

Find the global extrema in the interval $[0,4]$.

$$\begin{gathered} f\left(0\right)=0^{\frac{3}{2}}\left(3\right(0)-5) \\ =0 \\ f\left(4\right)=4^{\frac{3}{2}}\left(3\right(4)-5) \\ =56 \end{gathered}$$

The global maximum is at $x=4 with f\left(4\right)=56$ and the global minimum is at $x=1 with f\left(1\right)=-2$.

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 $[0,4)$.

$$\begin{gathered} f\left(0\right)=0^{\frac{3}{2}}\left(3\right(0)-5) \\ =0 \\ \underset{x\to 4^{-}}{lim}f\left(x\right)=\underset{x\to 4^{-}}{lim}\left(x^{\frac{3}{2}}\right(3x-5\left)\right) \\ =56 \end{gathered}$$

The global minimum is at $x=1 with f\left(1\right)=-2$.

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 $[0,1)$.

$$\begin{gathered} f\left(0\right)=0^{\frac{3}{2}}\left(3\right(0)-5) \\ =0 \\ \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}\left(x^{\frac{3}{2}}\right(3x-5\left)\right) \\ =-2 \end{gathered}$$

The global maximum is at $x=0 with f\left(0\right)=0$ and no global minimum.

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 $(0,1)$.

$$\begin{gathered} \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\left(x^{\frac{3}{2}}\right(3x-5\left)\right) \\ =0 \\ \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}\left(x^{\frac{3}{2}}\right(3x-5\left)\right) \\ =-2 \end{gathered}$$

There is no global maximum and global minimum.

Part (d) Step 2. Graph the function.

The graph of the function is: