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, 0.5\leq x\leq5
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.
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.
choose the correct graph of the function.

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 $-1 + x=0$, which gives $x = 1$. Also, $f^\prime(x)$ is undefined at $x = 0$, but $0$ is not in the interval $[0.5,5]$.

Step3: Evaluate the function at critical and end - points

Evaluate $f(x)$ at $x=0.5$, $x = 1$, and $x = 5$.
$f(0.5)=\frac{1}{0.5}+\ln(0.5)=2-\ln 2$.
$f(1)=\frac{1}{1}+\ln(1)=1$.
$f(5)=\frac{1}{5}+\ln(5)$.
Since $2-\ln 2\approx2 - 0.693 = 1.307$, $\frac{1}{5}+\ln(5)\approx0.2+1.609 = 1.809$.

Answer:

The absolute maximum value is $\frac{1}{5}+\ln(5)$ occurs at $x = 5$.
The absolute minimum value is $1$ occurs at $x = 1$.