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 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 the function

The function is \(f(x)=\frac{1}{x}+\ln x\), \(0.6\leq x\leq3\).
Using the power rule \((x^n)^\prime = nx^{n - 1}\) and \((\ln x)^\prime=\frac{1}{x}\), the derivative \(f^\prime(x)=-\frac{1}{x^{2}}+\frac{1}{x}=\frac{- 1 + x}{x^{2}}\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(\frac{-1 + x}{x^{2}}=0\).
Since \(x^{2}\gt0\) for \(x\in[0.6,3]\) (except \(x = 0\) which is not in our domain), then \(-1+x=0\), and \(x = 1\).

Step3: Evaluate the function at the critical point and endpoints

• Evaluate \(f(x)\) at \(x=0.6\): \(f(0.6)=\frac{1}{0.6}+\ln(0.6)=\frac{5}{3}+\ln(0.6)\approx\frac{5}{3}-0.5108\approx1.1559\).
• 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:

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