Q. 13 - 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.

$$\begin{gathered} f\left(x\right)=-12x+6x^2+4x^3-3x^4 \\ \left(a\right) [-1,1] \\ \left(b\right) (-1,1) \\ \left(c\right) (-3,0] \\ \left(d\right) [0,3] \end{gathered}$$

Step-by-Step Solution

Answer

(a) The global maximum of the function $f\left(x\right)=-12x+6x^2+4x^3-3x^4$ on $[-1,1]$ is $x=-1$ and at the values $f(-1)=11$ and global minimum at and at the values.

(b) The function has no global maximum and no minimum at $(-1,1)$.

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

(d) The global maximum of the function on $[0,3]$ is $x=0$ and at the values $f\left(0\right)=0$  and global minimum at $x=3$ and at the values $f\left(3\right)=-117$.

Part (a) Step 1. Given Information.

The function:

$$\begin{gathered} f\left(x\right)=-12x+6x^2+4x^3-3x^4 \\ [-1,1] \end{gathered}$$

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

The critical points are given by,

$$\begin{gathered} f\left(x\right)=-12x+6x^2+4x^3-3x^4 \\ f\text{'}\left(x\right)=-12x^3+12x^2+12x-12 \\ f\text{'}\left(x\right)=0 \\ -12x^3+12x^2+12x-12=0 \\ -12x(x^2-x-1)-12=0 \\ x=-1,1 \end{gathered}$$

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

The critical points can tested as:

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=-36x^2+24x+12 \\ f\text{'}\text{'}(-1)=-36{(-1)}^2+24(-1)+12 \\ =-48<0 \\ f\text{'}\text{'}\left(1\right)=-36{\left(1\right)}^2+24\left(1\right)+12 \\ =0 \end{gathered}$$

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

The height of the local extrema is,

$$\begin{gathered} f(-1)=-12(-1)+6{(-1)}^2+4{(-1)}^3-3{(-1)}^4 \\ =11 \\ f\left(1\right)=-12\left(1\right)+6{\left(1\right)}^2+4{\left(1\right)}^3-3{\left(1\right)}^4 \\ =-5 \end{gathered}$$

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

Find the global extrema in the interval $[-1,1]$

$$\begin{gathered} f(-1)=-12(-1)+6{(-1)}^2+4{(-1)}^3-3{(-1)}^4 \\ =11 \\ f\left(1\right)=-12\left(1\right)+6{\left(1\right)}^2+4{\left(1\right)}^3-3{\left(1\right)}^4 \\ =-5 \end{gathered}$$

The global maximum is at $x=-1 with f(-1)=11$ and the global minimum is at $x=1 with f\left(1\right)=-5$.

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 $(-1,1)$

$$\begin{gathered} f\left(x\right)=\underset{x\to 1^{+}}{lim}(-12x+6x^2+4x^3-3x^4) \\ =11 \\ f\left(x\right)=\underset{x\to 1^{-}}{lim(}-12x+6x^2+4x^3-3x^4) \\ =-5 \end{gathered}$$

The function has no global maximum and no 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 $(-3,0]$

$$\begin{gathered} f\left(x\right)=\underset{x\to -3^{+}}{lim}(-12x+6x^2+4x^3-3x^4) \\ =-207 \\ f\left(0\right)=(-12(0)+6{\left(0\right)}^2+4{\left(0\right)}^3-3{\left(0\right)}^4) \\ =0 \end{gathered}$$

The global maximum is at $x=0 with values f\left(0\right)=0$and there is 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 $[0,3]$

$$\begin{gathered} f\left(0\right)=(-12(0)+6{\left(0\right)}^2+4{\left(0\right)}^3-3{\left(0\right)}^4) \\ =0 \\ f\left(3\right)=(-12(3)+6{\left(3\right)}^2+4{\left(3\right)}^3-3{\left(3\right)}^4) \\ =-117 \end{gathered}$$

The global maximum is $x=0$ at $f\left(0\right)=0$ and the global minimum is $x=3$ at $f\left(3\right)=-117$.

Part (d) Step 2. Graph the function.

The graph of the function is: