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
(type an ordered pair.)

Explanation:

Step1: Find the derivative

Differentiate $f(x)=-x^{3}-9x^{2}-5$ using power - rule. The derivative $f^\prime(x)=-3x^{2}-18x=-3x(x + 6)$.

Step2: Find critical points

Set $f^\prime(x)=0$. So, $-3x(x + 6)=0$. Solving gives $x = 0$ and $x=-6$.

Step3: Set up intervals

The critical points divide the number - line into intervals $(-\infty,-6)$, $(-6,0)$ and $(0,\infty)$.

Step4: Test intervals

For $x\in(-\infty,-6)$, let $x=-7$. Then $f^\prime(-7)=-3\times(-7)\times(-7 + 6)=-21<0$.
For $x\in(-6,0)$, let $x=-1$. Then $f^\prime(-1)=-3\times(-1)\times(-1 + 6)=15>0$.
For $x\in(0,\infty)$, let $x = 1$. Then $f^\prime(1)=-3\times1\times(1 + 6)=-21<0$.

Step5: Determine relative extrema

Since $f(x)$ is decreasing on $(-\infty,-6)$ and increasing on $(-6,0)$, $x=-6$ is a relative minimum.
$f(-6)=-(-6)^{3}-9\times(-6)^{2}-5=216-324 - 5=-113$.
Since $f(x)$ is increasing on $(-6,0)$ and decreasing on $(0,\infty)$, $x = 0$ is a relative maximum.

Answer:

$(-6,-113)$