Chapter 4, Problem 3RE

To determine

To find: The local maximum and minimum and the absolute maximum and absolute minimum of the function $f\left(x\right)=\frac{3x-4}{x^2+1}$ on the interval $\left[-2,2\right]$.

Answer to Problem 3RE

The local minimum occurs at $x=-\frac{1}{3}$.

The absolute maximum occurs at $x=2$.

The absolute minimum is $x=-\frac{1}{3}$.

Explanation of Solution

Given:

The function is, $f\left(x\right)=\frac{3x-4}{x^2+1}$ and the interval is, $\left[-2,2\right]$.

Calculation:

Obtain the first derivative of the given function.

$\begin{array}{c}f\text{'}\left(x\right)=\frac{\left(x^2+1\right)\left(3\right)-\left(3x-4\right)\left(2x\right)}{{\left(x^2+1\right)}^2}\\ =\frac{3x^2+3-6x^2+8x}{{\left(x^2+1\right)}^2}\\ =\frac{-3x^2+8x+3}{{\left(x^2+1\right)}^2}\end{array}$

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

$\begin{array}{l}\frac{-3x^2+8x+3}{{\left(x^2+1\right)}^2}=0\\ -3x^2+8x+3=0\end{array}$

Apply quadratic formula and find the value of x.

$\begin{array}{c}x=\frac{-8\pm \sqrt{8^2-4\left(-3\right)\left(3\right)}}{2\left(-3\right)}\\ =\frac{-8\pm \sqrt{64+36}}{-6}\\ =\frac{8\pm 10}{6}\\ =\frac{18}{6}\text{ or }\frac{-2}{6}\end{array}$

Thus, the value of $x=-\frac{1}{3},3$.

Thus, the critical numbers are $x=-\frac{1}{3},3$ in which 3 does not lie on the given interval $\left[-2,2\right]$.

Apply the extreme values of the given interval and the critical number $x=-\frac{1}{3}$ in $f\left(x\right)$.

Substitute $x=-2$ in $f\left(x\right)$,

$\begin{array}{c}f\left(-2\right)=\frac{3\left(-2\right)-4}{{\left(-2\right)}^2+1}\\ =\frac{-6-4}{4+1}\\ =\frac{-10}{5}\\ =-2\end{array}$

Substitute $x=-\frac{1}{3}$ in $f\left(x\right)$,

$\begin{array}{c}f\left(-\frac{1}{3}\right)=\frac{3\left(-\frac{1}{3}\right)-4}{{\left(-\frac{1}{3}\right)}^2+1}\\ =\frac{-1-4}{\frac{1}{9}+1}\\ =\frac{-5}{\frac{10}{9}}\\ =-\frac{9}{2}\end{array}$

Substitute $x=2$ in $f\left(x\right)$,

$\begin{array}{c}f\left(2\right)=\frac{3\left(2\right)-4}{{\left(2\right)}^2+1}\\ =\frac{6-4}{4+1}\\ =\frac{2}{5}\end{array}$

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 approximately $\frac{2}{5}$ and the absolute minimum of $f\left(x\right)$ is −4.5.

Therefore, the absolute maximum of $f\left(x\right)$ occurs at $x=2$ and the absolute minimum of $f\left(x\right)$ is occurs at $x=-\frac{1}{3}$.

To find the local maximum and minimum obtain the second derivative of the function.

$\begin{array}{c}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(\frac{-3x^2+8x+3}{{\left(x^2+1\right)}^2}\right)\\ =\frac{-6x+8}{2\left(x^2+1\right)\left(2x\right)}\\ =\frac{2\left(-3x+4\right)}{4x\left(x^2+1\right)}\\ =\frac{-3x+4}{2x\left(x^2+1\right)}\end{array}$.

Substitute the critical numbers in the second derivative and obtain the local maximum and minimum as follows.

Substitute $x=-\frac{1}{3}$ in $f\text{'}\text{'}\left(x\right)$,

$\begin{array}{c}f\text{'}\text{'}\left(-\frac{1}{3}\right)=\frac{-3\left(\frac{-1}{3}\right)+4}{2\left(\frac{-1}{3}\right)\left({\left(\frac{-1}{3}\right)}^2+1\right)}\\ =\frac{1+4}{\left(\frac{-2}{3}\right)\left(\frac{1}{9}+1\right)}\\ =\frac{5}{\left(\frac{-2}{3}\right)\left(\frac{10}{9}\right)}\\ =-\frac{27}{4}<0\end{array}$

Thus, the local minimum occurs at $x=-\frac{1}{3}$.