Q. 49 - Calculus

Question

Use the Extreme Value Theorem to show that each function f has both a maximum and a minimum value on [a, b]. Then use a graphing utility to approximate values M and m in [a, b] at which f has a maximum and a minimum, respectively. You may assume that these functions are continuous everywhere. 

$f\left(x\right)=x^4-3x^2-2, [a,b]=[-2,2]$

Step-by-Step Solution

Answer

$$\begin{gathered} M=-2,2 \\ m=-1.44,1.44 \end{gathered}$$

Step 1. Given information.

We have been given a function and an interval as:

$f\left(x\right)=x^4-3x^2-2, [a,b]=[-2,2]$

We have to show that this function f has both a maximum and a minimum value on [a, b] using the Extreme Value Theorem. 

Also, we have to find approximate values M and m in [a, b] at which f has a maximum and a minimum, respectively, using a graphing utility.

Step 2. Apply the Extreme Value Theorem

$$\begin{gathered} \underset{x\to -2}{lim} f\left(x\right) \\ =\underset{x\to -2}{lim} \left(x^4-3x^2-2\right) \\ =(-2{)}^4-3(-2{)}^2-2 \\ =16-3\left(4\right)-2 \\ =14-12 \\ =2 \\ \underset{x\to 2}{lim} f\left(x\right) \\ =\underset{x\to 2}{lim} \left(x^4-3x^2-2\right) \\ =(2{)}^4-3(2{)}^2-2 \\ =16-3\left(4\right)-2 \\ =14-12 \\ =2 \end{gathered}$$

Step 3. Draw the graph of the given function

Step 4. Find M and m at which f has a maximum and a minimum

The maximum value of the function in the interval is $M=-2,2$.

The maximum value of the function in the interval is $m=-1.22,1.22$.