find the intervals on which f(x) is increasing and the intervals on which f(x) is decreasing. then sketch…

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 (type your answer using interval notation. use a comma to separate answers as needed.) b. the function is never increasing

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 (type your answer using interval notation. use a comma to separate answers as needed.) b. the function is never increasing

Answer

Explanation:

Step1: Find the derivative

Differentiate $f(x)=x^{4}-8x^{2}$ using power - rule. The derivative $f'(x)=4x^{3}-16x = 4x(x^{2}-4)=4x(x - 2)(x + 2)$.

Step2: Find the critical points

Set $f'(x)=0$. Then $4x(x - 2)(x + 2)=0$. Solving gives $x=-2,0,2$.

Step3: Test intervals

Consider the intervals $(-\infty,-2)$, $(-2,0)$, $(0,2)$ and $(2,\infty)$. For $x\in(-\infty,-2)$, let $x=-3$. Then $f'(-3)=4\times(-3)\times((-3)^{2}-4)=4\times(-3)\times5=-60<0$, so $f(x)$ is decreasing on $(-\infty,-2)$. For $x\in(-2,0)$, let $x = - 1$. Then $f'(-1)=4\times(-1)\times((-1)^{2}-4)=4\times(-1)\times(-3)=12>0$, so $f(x)$ is increasing on $(-2,0)$. For $x\in(0,2)$, let $x = 1$. Then $f'(1)=4\times1\times(1^{2}-4)=4\times1\times(-3)=-12<0$, so $f(x)$ is decreasing on $(0,2)$. For $x\in(2,\infty)$, let $x = 3$. Then $f'(3)=4\times3\times(3^{2}-4)=4\times3\times5 = 60>0$, so $f(x)$ is increasing on $(2,\infty)$.

Answer:

A. The function is increasing on $(-2,0),(2,\infty)$