Chapter 3, 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 intervals $f\left(x\right)=x^2+5x$.

Answer to Problem 1RE

Solution:

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

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:

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

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

Calculation:

Further differentiating the givenfunction,

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

To obtain the critical points in the given interval, solve the first derivative to zero.

$\begin{array}{c}2x+5=0\\ 2x=-5\\ x=\frac{-5}{2}\end{array}$

Calculating the value of the function at these critical points.

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

We compute the value of the function at the left and right extreme points:-

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

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