Question

the graph of the following function has one relative maximum and one relative minimum point. find these points using the first - derivative test

f(x)=-x^3 - 9x^2 - 5

the relative minimum is (-6, - 113)
(type an ordered pair.)

the relative maximum is
(type an ordered pair.)

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 $-60$.
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:

$(0,-5)$