Question

find the x-values of all points where the function has any relative extrema. find the value(s) of any relative extrema.
f(x)=-x³ - 3x² + 9x + 5
select the correct choice below and, if necessary, fill in any answer boxes within your choice.
a. there are no relative maxima. the function has a relative minimum of at x=
(use a comma to separate answers as needed.)
b. there are no relative minima. the function has a relative maximum of at x=
(use a comma to separate answers as needed.)
c. the function has a relative minimum of -22 at x=-3 and a relative maximum of 10 at x=1
(use a comma to separate answers as needed.)
d. there are no relative extrema.

Explanation:

Step1: Find the first derivative

Given \( f(x)=-x^{3}-3x^{2}+9x + 5\).
Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we have \(f^\prime(x)=-3x^{2}-6x + 9\).

Step2: Set the first derivative equal to zero

Set \(f^\prime(x)=0\), so \(-3x^{2}-6x + 9 = 0\).
Divide through by \(-3\) to get \(x^{2}+2x - 3=0\).
Factor the quadratic equation: \((x + 3)(x - 1)=0\).
Solve for \(x\): \(x=-3\) or \(x = 1\).

Step3: Find the second derivative

Differentiate \(f^\prime(x)=-3x^{2}-6x + 9\) to get \(f^{\prime\prime}(x)=-6x-6\).

Step4: Use the second - derivative test

For \(x=-3\):
Substitute \(x=-3\) into \(f^{\prime\prime}(x)\), \(f^{\prime\prime}(-3)=-6\times(-3)-6=18 - 6=12>0\).
Since \(f^{\prime\prime}(-3)>0\), the function has a relative minimum at \(x=-3\).
\(f(-3)=-(-3)^{3}-3(-3)^{2}+9(-3)+5=27-27-27 + 5=-22\).
For \(x = 1\):
Substitute \(x = 1\) into \(f^{\prime\prime}(x)\), \(f^{\prime\prime}(1)=-6\times1-6=-12<0\).
Since \(f^{\prime\prime}(1)<0\), the function has a relative maximum at \(x = 1\).
\(f(1)=-(1)^{3}-3(1)^{2}+9(1)+5=-1-3 + 9+5=10\).

Answer:

C. The function has a relative minimum of \(-22\) at \(x=-3\) and a relative maximum of \(10\) at \(x = 1\).