find the intervals on which f is increasing and the intervals on which it is decreasing. f(x)=-x^4/4 - 3x^3…

find the intervals on which f is increasing and the intervals on which it is decreasing. f(x)=-x^4/4 - 3x^3 - 9x^2 select the correct choice and, if necessary, fill in the answer box(es) to complete your choice. a. the function is decreasing on the open interval(s) . the function is never increasing. (simplify your answers. type your answers in interval notation. use a comma to separate answers as needed.) b. 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.) c. the function is increasing on the open interval(s) . the function is never decreasing. (simplify your answers. type your answers 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 the intervals on which it is decreasing. f(x)=-x^4/4 - 3x^3 - 9x^2 select the correct choice and, if necessary, fill in the answer box(es) to complete your choice. a. the function is decreasing on the open interval(s) . the function is never increasing. (simplify your answers. type your answers in interval notation. use a comma to separate answers as needed.) b. 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.) c. the function is increasing on the open interval(s) . the function is never decreasing. (simplify your answers. type your answers 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

Differentiate $f(x)=-\frac{x^{4}}{4}-3x^{3}-9x^{2}$ using power - rule. $f'(x)=-x^{3}-9x^{2}-18x=-x(x^{2}+9x + 18)=-x(x + 3)(x+6)$.

Step2: Find the critical points

Set $f'(x) = 0$. Then $-x(x + 3)(x + 6)=0$. The solutions are $x=-6,x=-3,x = 0$.

Step3: Test the intervals

Consider the intervals $(-\infty,-6),(-6,-3),(-3,0),(0,\infty)$. For $x\in(-\infty,-6)$, let $x=-7$. Then $f'(-7)=-(-7)(-7 + 3)(-7+6)=-(-7)(-4)(-1)=28>0$, so $f(x)$ is increasing on $(-\infty,-6)$. For $x\in(-6,-3)$, let $x=-4$. Then $f'(-4)=-(-4)(-4 + 3)(-4+6)=-( - 4)(-1)(2)=-8<0$, so $f(x)$ is decreasing on $(-6,-3)$. For $x\in(-3,0)$, let $x=-1$. Then $f'(-1)=-(-1)(-1 + 3)(-1+6)=-( - 1)(2)(5)=10>0$, so $f(x)$ is increasing on $(-3,0)$. For $x\in(0,\infty)$, let $x = 1$. Then $f'(1)=-(1)(1 + 3)(1+6)=-(1)(4)(7)=-28<0$, so $f(x)$ is decreasing on $(0,\infty)$.

Answer:

B. The function is increasing on the open interval(s) $(-\infty,-6),(-3,0)$ and decreasing on the open interval(s) $(-6,-3),(0,\infty)$