Question

find the intervals on which f is increasing and decreasing. f(x)=x² - 8 ln x select the correct choice below and, if necessary, fill in the answer box(es) within your choice. o 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.) o b. 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) o c. 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.) o d. the function is never increasing or decreasing

Explanation:

Step1: Find the domain and derivative

The domain of \(y = f(x)=x^{2}-8\ln x\) is \(x>0\).
The derivative \(f^{\prime}(x)=\frac{d}{dx}(x^{2}-8\ln x)\).
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\) and \(\frac{d}{dx}(\ln x)=\frac{1}{x}\), we get \(f^{\prime}(x)=2x-\frac{8}{x}=\frac{2x^{2}-8}{x}=\frac{2(x^{2}-4)}{x}=\frac{2(x - 2)(x + 2)}{x}\).

Step2: Find the critical points

Set \(f^{\prime}(x)=0\), so \(\frac{2(x - 2)(x + 2)}{x}=0\) (since \(x>0\)).
The critical point is \(x = 2\) (because \(x=-2\) is not in the domain \(x>0\)).

Step3: Test the intervals

• For the interval \((0,2)\), let \(x = 1\). Then \(f^{\prime}(1)=\frac{2(1 - 2)(1+ 2)}{1}=\frac{2\times(-1)\times3}{1}=-6<0\). So \(f(x)\) is decreasing on \((0,2)\).
• For the interval \((2,\infty)\), let \(x = 3\). Then \(f^{\prime}(3)=\frac{2(3 - 2)(3 + 2)}{3}=\frac{2\times1\times5}{3}=\frac{10}{3}>0\). So \(f(x)\) is increasing on \((2,\infty)\).

Answer:

A. The function is increasing on the open interval(s) \((2,\infty)\) and decreasing on the open interval(s) \((0,2)\)