Q. 51 - 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. 

$f\left(x\right)=x^4-3x^2-2, [a,b]=[-1,1]$

Step-by-Step Solution

Answer

$$\begin{gathered} M=0 \\ m=-1,1 \end{gathered}$$

Step 1. Given information.

We have been given a function and an interval as:

$f\left(x\right)=x^4-3x^2-2, [a,b]=[-1,1]$

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

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

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 value of the function is maximum in the interval at $x=0$.

The value of the function is minimum in the interval at $x=-1,1$.