Q. 53 - 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. 

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

Step-by-Step Solution

Answer

$$\begin{gathered} M=0,2 \\ m=1.33 \end{gathered}$$

Step 1. Given information.

We have been given a function and an interval as:

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

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

$$\begin{gathered} \underset{x\to 0}{lim} f\left(x\right) \\ =\underset{x\to 0}{lim} \left(3-2x^2+x^3\right) \\ =3-2\left(0^2\right)+0^3 \\ =3-0+0 \\ =3 \\ \underset{x\to 2}{lim} f\left(x\right) \\ =\underset{x\to 2}{lim} \left(3-2x^2+x^3\right) \\ =3-2\left(2^2\right)+2^3 \\ =3-2\left(4\right)+8 \\ =3-8+8 \\ =3 \end{gathered}$$

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 $x=0, 2$.

The value of the function is minimum in the interval at $x=1.33$.