Chapter 4, Problem 1RE

Finding Extrema on a Closed Interval In Exercises 1-8, find the absolute extrema of the function on the closed interval.

$f(x)=x^2+5x,[-4,0]$

To determine

To calculate: The absolute extrema for the function $f\left(x\right)=x^2+5x$ in the provided intervals $f\left(x\right)=x^2+5x$.

Answer to Problem 1RE

Solution:

The maximum is $\underset{\_}{\left(0,0\right)}$ and the minimum is $\underset{\_}{\left(\frac{-5}{2},\frac{-25}{4}\right)}$.

Explanation of Solution

Given:

The function $f\left(x\right)=x^2+5x$ in the interval $f\left(x\right)=x^2+5x$.

Formula used:

A point c is said to be the critical point of the function f if $f\text{'}\left(c\right)=0$.

The extrema of the provided function f would exist at either the critical point or the end points of the closed interval.

Calculation:

First differentiate the provided function,

$f\text{'}\left(x\right)=2x+5$

Equate the first derivative to zero to obtain the critical points in the provided interval.

$\begin{array}{ccc}\hfill 2x+5& =& 0\hfill \\ \hfill 2x& =& -5\hfill \\ \hfill x& =& \frac{-5}{2}\hfill \end{array}$

Now calculate the value of the function at these critical points.

$\begin{array}{ccc}\hfill f\left(\frac{-5}{2}\right)& =& {\left(\frac{-5}{2}\right)}^2+5\left(\frac{-5}{2}\right)\hfill \\ \hfill & =& \frac{25}{4}-\frac{25}{2}\hfill \\ \hfill & =& \frac{-25}{4}\hfill \end{array}$

Now the value of the function at the left and right extreme points can be computed as:

$\begin{array}{ccc}\hfill f\left(0\right)& =& {\left(0\right)}^2+5\left(0\right)\hfill \\ \hfill & =& 0\hfill \\ \hfill f\left(-4\right)& =& {\left(-4\right)}^2+5\left(-4\right)\hfill \\ \hfill & =& -4\hfill \end{array}$

Hence, the maximum is $\left(0,0\right)$ and the minimum is $\left(\frac{-5}{2},\frac{-25}{4}\right)$.