Q. 12 - 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. 

$$\begin{gathered} f\left(x\right)=3x^4+4x^3-36x^2 on the interval \\ \left(a\right) [-5,5] \\ \left(b\right) [-2,2] \\ \left(c\right) (-3,1] \\ \left(d\right) (-1,3) \end{gathered}$$

Step-by-Step Solution

Answer

(a) The global maximum of the function $f\left(x\right)=3x^4+4x^3-36x^2$ on $[-5,5]$ is $x=0$ and at the values $f\left(0\right)=0$ and global minimum at $x=-3,2$ and at the values$f(-3)=-189, f\left(2\right)=-64$.

(b) The global maximum of the function $f\left(x\right)=3x^4+4x^3-36x^2$ on $[-2,2]$ is $x=0$ and at the values $f\left(0\right)=0$ and global minimum at  $x=-2,2$and at the values $f\left(2\right)=-64$ 

(c) The global maximum of the function $f\left(x\right)=3x^4+4x^3-36x^2$ on $(-3.1]$ is$x=0$  and at the values $f\left(0\right)=0$ and global minimum at  $x=1$and at the values $f\left(1\right)=-29$

(d) The global maximum of the function $f\left(x\right)=3x^4+4x^3-36x^2$ on $(-1,3)$ is $x=0$ and at the values $f\left(0\right)=0$ and global minimum at $x=1$ and at the values $f\left(1\right)=-29$

Part (a) Step 1. Given Information.

The function: 

$$\begin{gathered} f\left(x\right)=3x^4+4x^3-36x^2 \\ [-5,5] \end{gathered}$$

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

The critical points are given by, 

$$\begin{gathered} f\left(x\right)=3x^4+4x^3-36x^2 \\ f\text{'}\left(x\right)=12x^3+12x^2-72x \\ {x}^2+x-6=0 \\ x=0,-3,2 \end{gathered}$$

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

The critical points can tested as: 

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=36x^2+24x-72 \\ f\text{'}\text{'}\left(0\right)=-72<0 \\ f\text{'}\text{'}(-3)=180>0 \\ f\left(2\right)=120>0 \end{gathered}$$

So the function has a local maximum at

The height of the local extrema is,

$$\begin{gathered} f\left(0\right)=0 \\ f(-3)=-189 \\ f\left(2\right)=-64 \end{gathered}$$

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

Find the global extrema in the interval $[-5,5]$

$$\begin{gathered} f(-5)=3{(-5)}^4+4{(-5)}^3-36{(-5)}^2 \\ =475 \\ f\left(5\right)=3{\left(5\right)}^4+4{\left(5\right)}^3-36{\left(5\right)}^2 \\ =1475 \end{gathered}$$

The global maximum is at $x=0 with f\left(0\right)=0$ and the global minimum is at $x=-3,2 with f(-3)=-189, f\left(2\right)=-64$

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 $[-2,2]$.

$$\begin{gathered} f(-2)=3{(-2)}^4+4{(-2)}^3-36{(-2)}^2 \\ =-128 \\ f\left(2\right)=3{\left(2\right)}^4+4{\left(2\right)}^3-36{\left(2\right)}^2 \\ =-64 \end{gathered}$$

The global maximum is at $x=0 with f\left(0\right)=0$ and the global minimum is at  $x=-2,2 with f\left(2\right)=-64$

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 $(-3,1]$

$$\begin{gathered} \underset{x\to -3^{+}}{lim} f\left(x\right)=\underset{x\to -3^{+}}{lim}3x^4+4x^3-36x^2 \\ =-189 \\ f\left(1\right)=3{\left(1\right)}^4+4{\left(1\right)}^3-36{\left(1\right)}^2 \\ =-29 \end{gathered}$$

The global maximum is at $x=0 with f\left(0\right)=0$ and the global minimum is at $x=1 with f\left(1\right)=-29$

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 $(-1,3)$

$$\begin{gathered} \underset{x\to -1^{+}}{lim} f\left(x\right)=\underset{x\to -1^{+}}{lim}3x^4+4x^3-36x^2 \\ =-37 \\ \underset{x\to 3^{-}}{lim} f\left(x\right)=\underset{x\to 3^{-}}{lim}3x^4+4x^3-36x^2 \\ =-9 \end{gathered}$$

The global maximum is at $x=0 ith f\left(0\right)=0$ and the global minimum is at $x=1$ with $f\left(1\right)=-29$

Part (d) Step 2. Graph the function.

The graph of the function is: