Chapter 5.1, Problem 11E

To determine

To find: theextreme values of the function on the interval and where they occur by using analytic method and find critical points that are not stationary points.

Answer to Problem 11E

the maximum value is $\frac{1}{4}+\ln 4$ at $x=4$ and minimum value is $1$ at $x=1$ and local minimum is $\left(0.5,2-\ln 2\right)$ and function have no critical points.

Explanation of Solution

Given information:

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

Calculation: to find the extreme values of function, first find its derivative,

  $\begin{array}{l}f\left(x\right)=\frac{1}{x}+\ln x\\ f\text{'}\left(x\right)=-\frac{1}{x^2}+\frac{1}{x}\text{ [differentiate]}\end{array}$

There is no points where $f\text{'}\left(x\right)=0$ so the function does not have any critical points, and the point $x=0.5$ the $f\text{'}\left(x\right)<0$ so at point $x=0.5$ ,the function have local minimum value and point $x=4$ , $f\text{'}\left(x\right)>0$ so at the point $x=4$ function have maximum value.

Now substitute in function,

  $\begin{array}{l}f\left(x\right)=\frac{1}{x}+\ln x\\ f\left(0.5\right)=2-\ln \left(2\right)\text{ [substitute }x=2\text{ and simplify]}\\ f\left(4\right)=\frac{1}{4}+\ln 4\text{ [substitute }x=4\text{ ]}\\ f\left(1\right)=1\text{ [substitute }x=1\text{ and simplify]}\end{array}$

Thus, the maximum value is $\frac{1}{4}+\ln 4$ at $x=4$ and minimum value is $1$ at $x=1$ and local minimum is $\left(0.5,2-\ln 2\right)$ .