Question

find the absolute maximum and minimum values of the following function on the given interval. then graph the function.
f(x) = \frac{1}{x}+\ln x, 0.6\leq x\leq8

a. the absolute maximum value \frac{1}{8}+3\ln(2) occurs at x = 8
(type exact answers. use a comma to separate answers as needed.)
b. there is no absolute maximum.

find the absolute minimum value. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice

a. the absolute minimum value occurs at x =
(type exact answers. use a comma to separate answers as needed.)
b. there is no absolute minimum.

Explanation:

Step1: Find the derivative

The derivative of $f(x)=\frac{1}{x}+\ln x$ is $f^\prime(x)=-\frac{1}{x^{2}}+\frac{1}{x}=\frac{- 1 + x}{x^{2}}$.

Step2: Find critical points

Set $f^\prime(x) = 0$, so $\frac{-1 + x}{x^{2}}=0$. Then $-1 + x=0$, and we get $x = 1$. Also, $f^\prime(x)$ is undefined at $x = 0$, but $0$ is not in the interval $[0.6,8]$.

Step3: Evaluate the function at critical - point and endpoints

Evaluate $f(x)$ at $x = 0.6$, $x=1$, and $x = 8$.
$f(0.6)=\frac{1}{0.6}+\ln(0.6)=\frac{5}{3}+\ln(0.6)$;
$f(1)=\frac{1}{1}+\ln(1)=1$;
$f(8)=\frac{1}{8}+\ln(8)=\frac{1}{8}+3\ln(2)$.

Step4: Compare the values

$\frac{5}{3}+\ln(0.6)\approx1.67-0.51 = 1.16$, $1$, $\frac{1}{8}+3\ln(2)\approx0.125 + 2.079=2.204$.

Answer:

A. The absolute minimum value $1$ occurs at $x = 1$.