Q. 54 - Calculus

Question

Use the Extreme Value Theorem to show that each function f in Exercises 49–54 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.

$$\begin{gathered} f\left(x\right)=3-2x^2+x^3; \\ [a,b]=[-1,1] \end{gathered}$$

Step-by-Step Solution

Answer

The function $f\left(x\right)=3-2x^2+x^3$ has both maximum and minimum value on the given interval. The maximum value M is $M=2$ and the minimum value m is $m=0$.

And by using graphing utility, approximation of the given function is:

Step 1. Given Information.

The function:

$$\begin{gathered} f\left(x\right)=3-2x^2+x^3; \\ [a,b]=[-1,1] \end{gathered}$$

Step 2. Find minimum value.

Substitute the lower interval in the given function,

$$\begin{gathered} f(-1)={(-1)}^3-2{(-1)}^2+3 \\ m=-1-2+3 \\ m=0 \end{gathered}$$

Step 3. Find the maximum value.

Substitute the upper interval in the given function,

$$\begin{gathered} f\left(1\right)=3-2{\left(1\right)}^2+{\left(1\right)}^3 \\ M=3-2+1 \\ M=2 \end{gathered}$$

Step 4. About maximum and minimum points

The maximum value of the function occurs at $M=2$ and the minimum occurs at $m=0$

Step 5. Graph the function

Graph the function,

From the graph, the function has a maximum value and the minimum value which occurs in the interval $[-1,1]$