Chapter 4.1, Problem 11E

To determine

To find : extreme values of the function on the interval using the Candidates test.

Answer to Problem 11E

Maximum occur at $\left(0.5,1.307\right)$ and $\left(4,1.636\right)$ .

Minimum occur at $\left(1,1\right)$ .

Explanation of Solution

Given information :

  $f\left(x\right)=\frac{1}{x}+\ln x$ , $0.5\le x\le 4$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Differentiate $f\left(x\right)=\frac{1}{x}+\ln x$ with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{1}{x}+\ln x\right)\\ =\frac{d}{dx}\left(\frac{1}{x}\right)+\frac{d}{dx}\left(\ln x\right)\\ =-\frac{1}{x^2}+\frac{1}{x}\\ =\frac{-1+x}{x^2}\end{array}$

Take $\frac{dy}{dx}=0$ .

  $\begin{array}{c}\frac{-1+x}{x^2}=0\\ -1+x=0\left[x^2\ne 0\right]\\ x=1\end{array}$

So, there is only one critical points at $x=1$ .

Plug the critical point in the function.

  $\begin{array}{c}f\left(1\right)=\frac{1}{1}+\ln 1\\ =1+0\\ =1\end{array}$

Plug the end points in the function.

  $\begin{array}{c}f\left(0.5\right)=\frac{1}{0.5}+\ln 0.5\\ =2+\ln 0.5\\ =1.307\end{array}$

And

  $\begin{array}{c}f\left(4\right)=\frac{1}{4}+\ln 4\\ =0.25+\ln 4\\ =1.636\end{array}$

Maximum occur at $\left(0.5,1.307\right)$ and $\left(4,1.636\right)$ .

Minimum occur at $\left(1,1\right)$ .