Chapter 4, Problem 2RCC

(a)

To determine

To explain: The Extreme value theorem.

Explanation of Solution

The extreme value theorem is stated as follow.

Extreme Value Theorem:

“If f is a continuous function on $\left[a,b\right]$ then f attains its absolute maximum value $f\left(c\right)$ and absolute minimum value $f\left(d\right)$ at some points c and d in the interval $\left[a,b\right]$”.

From the following Figure 1, the meaning of extreme value theorem can be easily understood.

In the Figure 1 the function f is defined on a closed interval $\left[a,b\right]$ and it attains its maximum value at $f\left(c\right)$ and minimum value at $f\left(d\right)$ where c and d are lies in the closed interval $\left[a,b\right]$.

(b)

To determine

To explain: The working of Closed interval method.

Explanation of Solution

Closed interval method is an algorithm that is used to find the absolute minimum and the absolute maximum of a function over a closed interval $\left[a,b\right]$.

It consists of following three steps.

Step 1: Find the values of function f at the critical points in $\left(a,b\right)$.

Step 2: Find the values of function f at the end points of the interval $\left[a,b\right]$.

Step 3: Largest of the values from step 1 and step 2 is the absolute maximum value and lowest one is absolute minimum value.

Apply the closed interval method to find the absolute maximum and minimum of a function as illustrated in following example.

Example:

The absolute maximum and absolute minimum of the function $f\left(x\right)=5+54x-2x^3$ over the interval $\left[0,4\right]$

Calculation:

Obtain the first derivative of the given function.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(5+54x-2x^3\right)\\ =\frac{d}{dx}\left(5\right)+\frac{d}{dx}\left(54x\right)-\frac{d}{dx}\left(2x^3\right)\\ =0+54\cdot (1)-2\cdot \left(3x^2\right)\\ =54-6x^2\end{array}$

Set $f^{\prime }\left(x\right)=0$ and obtain the critical numbers.

$\begin{array}{l}54-6x^2=0\\ x^2=9\\ x=3\text{ or }-3\end{array}$

Here, the critical number $x=3$ lies in the given interval $\left[0,4\right]$ while the other critical point $x=-3$ does not lie in $\left[0,4\right]$.

Step 1: Find the values of function f at the critical points in $\left(0,4\right)$.

Here the critical point is $x=3$. So, substitute $x=3$ in $f\left(x\right)$,

$\begin{array}{c}f\left(3\right)=5+54\left(3\right)-2{\left(3\right)}^3\\ =5+162-54\\ =113\end{array}$

Step 2: Apply the extreme values of the given interval in $f\left(x\right)$.

Substitute $x=0$ in $f\left(x\right)$,

$\begin{array}{c}f\left(0\right)=5+54\left(0\right)-2{\left(0\right)}^3\\ =5\end{array}$

Substitute $x=4$ in $f\left(x\right)$,

$\begin{array}{c}f\left(4\right)=5+54\left(4\right)-2{\left(4\right)}^3\\ =5+216-128\\ =93\end{array}$

Step 3:

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of $f\left(x\right)$ is 113 and the absolute minimum of $f\left(x\right)$ is 5.

Therefore, the absolute maximum value of $f\left(x\right)$ is $f\left(3\right)=113$ and the absolute minimum of $f\left(x\right)$ is $f\left(0\right)=5$.