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\leq4
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 of $f(x)$

$f(x)=\frac{1}{x}+\ln x=x^{-1}+\ln x$. Using the power - rule and the derivative of the natural logarithm, $f^\prime(x)=-x^{-2}+\frac{1}{x}=-\frac{1}{x^{2}}+\frac{1}{x}=\frac{- 1 + x}{x^{2}}$.

Step2: Find the critical points

Set $f^\prime(x) = 0$, so $\frac{-1 + x}{x^{2}}=0$. Since $x^{2}
eq0$ for $x\in[0.5,4]$, we solve $-1 + x = 0$, and we get $x = 1$. Also, $f^\prime(x)$ is undefined when $x = 0$, but $0
otin[0.5,4]$, so we ignore it.

Step3: Evaluate the function at the critical point and endpoints

Evaluate $f(x)$ at $x = 0.5$, $x = 1$, and $x = 4$.
$f(0.5)=\frac{1}{0.5}+\ln(0.5)=2-\ln2\approx2 - 0.693 = 1.307$.
$f(1)=\frac{1}{1}+\ln(1)=1+0 = 1$.
$f(4)=\frac{1}{4}+\ln(4)=\frac{1}{4}+2\ln2\approx0.25 + 2\times0.693=0.25+1.386 = 1.636$.

Answer:

A. The absolute maximum value $1.636$ occurs at $x = 4$.