find the intervals on which f is increasing and decreasing. f(x)=x² - 18 ln x select the correct choice…

find the intervals on which f is increasing and decreasing. f(x)=x² - 18 ln x select the correct choice below and, if necessary, fill in the answer box(es) within your choice. a. the function is increasing on the open interval(s) and decreasing on the open interval(s) (simplify your answers. type your answers in interval notation. use a comma to separate answers as needed.) b. the function is decreasing on the open interval(s) the function is never increasing. (simplify your answer. type your answer in interval notation. use a comma to separate answers as needed.) c. the function is increasing on the open interval(s) the function is never decreasing. (simplify your answer. type your answer in interval notation. use a comma to separate answers as needed.) d. the function is never increasing or decreasing.

find the intervals on which f is increasing and decreasing. f(x)=x² - 18 ln x select the correct choice below and, if necessary, fill in the answer box(es) within your choice. a. the function is increasing on the open interval(s) and decreasing on the open interval(s) (simplify your answers. type your answers in interval notation. use a comma to separate answers as needed.) b. the function is decreasing on the open interval(s) the function is never increasing. (simplify your answer. type your answer in interval notation. use a comma to separate answers as needed.) c. the function is increasing on the open interval(s) the function is never decreasing. (simplify your answer. type your answer in interval notation. use a comma to separate answers as needed.) d. the function is never increasing or decreasing.

Answer

Explanation:

Step1: Find the derivative

The derivative of $f(x)=x^{2}-18\ln x$ using the power - rule $(x^n)' = nx^{n - 1}$ and $(\ln x)'=\frac{1}{x}$ is $f'(x)=2x-\frac{18}{x}=\frac{2x^{2}-18}{x}=\frac{2(x^{2}-9)}{x}=\frac{2(x - 3)(x + 3)}{x}$, with the domain $x>0$ (since $\ln x$ is defined for $x>0$).

Step2: Find the critical points

Set $f'(x)=0$. So $\frac{2(x - 3)(x + 3)}{x}=0$. Since $x>0$, we consider $(x - 3)(x + 3)=0$. The positive root is $x = 3$ (we ignore $x=-3$ as it is not in the domain).

Step3: Test intervals

We test the intervals $(0,3)$ and $(3,\infty)$. For the interval $(0,3)$, let's choose $x = 1$. Then $f'(1)=\frac{2(1 - 3)(1+3)}{1}=\frac{2\times(-2)\times4}{1}=-16<0$. So $f(x)$ is decreasing on $(0,3)$. For the interval $(3,\infty)$, let's choose $x = 4$. Then $f'(4)=\frac{2(4 - 3)(4 + 3)}{4}=\frac{2\times1\times7}{4}=\frac{7}{2}>0$. So $f(x)$ is increasing on $(3,\infty)$.

Answer:

A. The function is increasing on the open interval $(3,\infty)$ and decreasing on the open interval $(0,3)$