Q. 11 - 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. 

$$\begin{gathered} f\left(x\right)=2x^3-3x^2-12x on the interval \\ \left(a\right) [-3,3] \\ \left(b\right) [0,3] \\ \left(c\right) (-1,2] \\ \left(d\right) (-2,1) \end{gathered}$$

Step-by-Step Solution

Answer

(a) The global maximum of the function $f\left(x\right)=2x^3-3x^2-12x$ on $[-3,3]$ is $x=-1$ and at the values $f(-1)=7$ and global minimum at $x=-3$ and at the values $f(-3)=-45$.

(b) The global maximum of the function $f\left(x\right)=2x^3-3x^2-12x$ on $[0,3]$ is $x=0$ and at the values $f\left(0\right)=0$ and global minimum at $x=2$ and at the values $f\left(2\right)=-20$

(c) There is no global maximum and global minimum at $x=2$ and at the values $f\left(2\right)=-20$ 

(d) The global maximum of the function $f\left(x\right)=2x^3-3x^2-12x$ on $(-2,1)$ is $x=-1 with f(-1)=7$ and there is no global minimum.

Part (a) Step 1. Given Information.

The function: 

$$\begin{gathered} f\left(x\right)=2x^3-3x^2-12x \\ [-3,3] \end{gathered}$$

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

The critical points are given by, 

$$\begin{gathered} f\left(x\right)=2x^3-3x^2-12x \\ f\text{'}\left(x\right)=6x^2-6x-12 \\ f\text{'}\left(x\right)=0 \\ x^2-x-2=0 \\ (x-2)(x+1)=0 \\ x=-1,2 \end{gathered}$$

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

The critical points can tested as: 

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=12x-6 \\ f\text{'}\text{'}(-1)=12(-1)-6 \\ =-24<0 \\ f\text{'}\text{'}\left(2\right)=12\left(2\right)-6 \\ =18>0 \end{gathered}$$

So the function has a local maximum at

The height of the local extrema is,

$$\begin{gathered} f(-1)=2{(-1)}^3-3{(-1)}^2-12(-1) \\ =7 \\ f\left(2\right)=2{\left(2\right)}^3-3{\left(2\right)}^2-12\left(2\right) \\ =-20 \end{gathered}$$

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

Find the global extrema in the interval $[-3,3]$,

$$\begin{gathered} f(-3)=2{(-3)}^3-3{(-3)}^2-12(-3) \\ =-45 \\ f\left(3\right)=2{\left(3\right)}^3-3{\left(3\right)}^2-12\left(3\right) \\ =-9 \end{gathered}$$

The global maximum is at $x=-1, f(-1)=7$ and the global minimum is at $x=-3, f(-3)=-45$

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 $[0,3]$.

$$\begin{gathered} f\left(0\right)=2{\left(0\right)}^3-3{\left(0\right)}^2-12\left(0\right) \\ =0 \\ f\left(3\right)=2{\left(3\right)}^3-3{\left(3\right)}^2-12\left(3\right) \\ =-9 \end{gathered}$$

The global maximum is at $x=0 with f\left(0\right)=0$ and the global minimum is at $x=2 with f\left(2\right)=-20$.

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 $(-1,2]$.

$$\begin{gathered} \underset{x\to -1^{+}}{lim}f\left(x\right)=\underset{x\to -1^{+}}{lim}2x^3-3x^2-12x \\ =7 \\ f\left(2\right)=2{\left(2\right)}^3-3{\left(2\right)}^2-12\left(2\right) \\ =-20 \end{gathered}$$

There is no global maximum and the global minimum is at $x=2 with values f\left(2\right)=-20$

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 $(-2,1)$
$$\begin{gathered} \underset{x\to -2^{+}}{lim}f\left(x\right)=\underset{x\to -2^{+}}{lim}2x^3-3x^2-12x \\ =-4 \\ \underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}2x^3-3x^2-12x \\ =-13 \end{gathered}$$

The global maximum is at $x=-1 with f(-1)=7$ and there is no global minimum.

Part (d) Step 2. Graph the function.

The graph of the function is: