the graph of the following function has one relative maximum and one relative minimum point. find these…

the graph of the following function has one relative maximum and one relative minimum point. find these points using the first - derivative test\n\nf(x)=-x^3 - 9x^2 - 5\n\nthe relative minimum is (-6, - 113)\n(type an ordered pair.)\n\nthe relative maximum is \n(type an ordered pair.)

the graph of the following function has one relative maximum and one relative minimum point. find these points using the first - derivative test\n\nf(x)=-x^3 - 9x^2 - 5\n\nthe relative minimum is (-6, - 113)\n(type an ordered pair.)\n\nthe relative maximum is \n(type an ordered pair.)

Answer

Explanation:

Step1: Find the derivative

$f'(x)=-3x^{2}-18x=-3x(x + 6)$

Step2: Find critical points

Set $f'(x)=0$, then $-3x(x + 6)=0$. Solving gives $x = 0$ and $x=-6$.

Step3: Use first - derivative test

Choose test points in the intervals $(-\infty,-6)$, $(-6,0)$ and $(0,\infty)$. For $x<-6$, let $x=-7$, $f'(-7)=-3\times(-7)\times(-7 + 6)=-21<0$. For $-6<x<0$, let $x=-1$, $f'(-1)=-3\times(-1)\times(-1 + 6)=15>0$. For $x>0$, let $x = 1$, $f'(1)=-3\times1\times(1 + 6)=-21<0$. Since the function changes from decreasing ($f'(x)<0$) to increasing ($f'(x)>0$) at $x=-6$ (which is the given relative minimum), and changes from increasing to decreasing at $x = 0$. To find the $y$-coordinate of the relative maximum, substitute $x = 0$ into $f(x)$: $f(0)=-0^{3}-9\times0^{2}-5=-5$.

Answer:

$(0,-5)$