Question

find the intervals on which f(x) is increasing and the intervals on which f(x) is decreasing. then sketch the graph. add horizontal tangent lines. f(x)=x^4 - 8x^2. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the function is increasing on (-2,0),(2,∞) (type your answer using interval notation. use a comma to separate answers as needed.) b. the function is never increasing. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the function is decreasing on (type your answer using interval notation. use a comma to separate answers as needed.) b. the function is never decreasing.

Explanation:

Step1: Find the derivative

Given $f(x)=x^{4}-8x^{2}$, using the power - rule $(x^n)' = nx^{n - 1}$, we have $f'(x)=4x^{3}-16x=4x(x^{2}-4)=4x(x - 2)(x + 2)$.

Step2: Find the critical points

Set $f'(x)=0$, then $4x(x - 2)(x + 2)=0$. The solutions are $x=-2,0,2$.

Step3: Test the intervals

We consider the intervals $(-\infty,-2)$, $(-2,0)$, $(0,2)$ and $(2,\infty)$.
For $x\in(-\infty,-2)$, let $x=-3$, then $f'(-3)=4\times(-3)\times(-3 - 2)\times(-3 + 2)=4\times(-3)\times(-5)\times(-1)=-60<0$, so the function is decreasing on $(-\infty,-2)$.
For $x\in(-2,0)$, let $x=-1$, then $f'(-1)=4\times(-1)\times(-1 - 2)\times(-1 + 2)=4\times(-1)\times(-3)\times1 = 12>0$, so the function is increasing on $(-2,0)$.
For $x\in(0,2)$, let $x = 1$, then $f'(1)=4\times1\times(1 - 2)\times(1 + 2)=4\times1\times(-1)\times3=-12<0$, so the function is decreasing on $(0,2)$.
For $x\in(2,\infty)$, let $x = 3$, then $f'(3)=4\times3\times(3 - 2)\times(3 + 2)=4\times3\times1\times5 = 60>0$, so the function is increasing on $(2,\infty)$.

Answer:

The function is decreasing on $(-\infty,-2),(0,2)$