Q. 50 - 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. 

$f\left(x\right)=x^4-3x^2-2, [a,b]=[0,2]$

Step-by-Step Solution

Answer

The extreme value function guarantees that the function will not attain global maximum or minimum values on the given interval.

Step 1. Given information.

We have been given a function and an interval as:

$f\left(x\right)=x^4-3x^2-2, [a,b]=[0,2]$

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

$$\begin{gathered} \underset{x\to 0}{lim} f\left(x\right) \\ =\underset{x\to 0}{lim} \left(x^4-3x^2-2\right) \\ =0^4-3(0{)}^2-2 \\ =0-3(0)-2 \\ =0-2 \\ =-2 \\ \underset{x\to 2}{lim} f\left(x\right) \\ =\underset{x\to 2}{lim} \left(x^4-3x^2-2\right) \\ =(2{)}^4-3(2{)}^2-2 \\ =16-3(4)-2 \\ =14-12 \\ =2 \end{gathered}$$

The extreme value function guarantees that the function will not attain global maximum or minimum values on the interval [0,2].