Question

find the absolute maximum and minimum values of the following function on the given interval. then graph the function.

$f(x)=\frac{1}{x}+\ln x, 0.6 \leq x \leq 3$

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 $\square$ occurs at $x=\square$
(type exact answers. use a comma to separate answers as needed)

b. there is no absolute maximum.

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=\frac{1}{x}+\ln x\), its derivative \(f^{\prime}(x)=-\frac{1}{x^{2}}+\frac{1}{x}=\frac{-1 + x}{x^{2}}\)

Step2: Find the critical points

Set \(f^{\prime}(x) = 0\), then \(\frac{-1 + x}{x^{2}}=0\). Since \(x^{2}>0\) for \(x\in[0.6,3]\) (except \(x = 0\) which is not in our domain), we solve \(-1+x=0\), getting \(x = 1\)

Step3: Evaluate the function at critical points and endpoints

• Evaluate \(f(x)\) at \(x=0.6\): \(f(0.6)=\frac{1}{0.6}+\ln(0.6)=\frac{5}{3}+\ln\frac{3}{5}\approx\frac{5}{3}- 0.5108\approx1.6667-0.5108 = 1.1559\)
• Evaluate \(f(x)\) at \(x = 1\): \(f(1)=\frac{1}{1}+\ln(1)=1+0 = 1\)
• Evaluate \(f(x)\) at \(x=3\): \(f(3)=\frac{1}{3}+\ln(3)\approx0.3333 + 1.0986=1.4319\)

Answer:

The absolute maximum value \(1.4319\) occurs at \(x = 3\)