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
find the absolute maximum value. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the absolute maximum value occurs at x=
(type exact answers. use a comma to separate answers as needed.)
b. there is no absolute maximum.

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 $\frac{-1 + x}{x^{2}}=0$. Since $x^{2}
eq0$ for the domain of our function ($x\in[0.6,8]$), then $-1 + x=0$, which gives $x = 1$.

Step3: Evaluate the function at critical - point and endpoints

Evaluate $f(x)$ at $x = 0.6$, $x = 1$, and $x = 8$.
When $x=0.6$, $f(0.6)=\frac{1}{0.6}+\ln(0.6)=\frac{10}{6}+\ln(0.6)=\frac{5}{3}+\ln(0.6)$.
When $x = 1$, $f(1)=\frac{1}{1}+\ln(1)=1+0 = 1$.
When $x = 8$, $f(8)=\frac{1}{8}+\ln(8)=\frac{1}{8}+3\ln(2)$.
We know that $\frac{5}{3}+\ln(0.6)\approx\frac{5}{3}- 0.5108\approx1.6667 - 0.5108 = 1.1559$, $1$, and $\frac{1}{8}+3\ln(2)\approx0.125+3\times0.6931=0.125 + 2.0793=2.2043$.

Answer:

A. The absolute maximum value $\frac{1}{8}+3\ln(2)$ occurs at $x = 8$.