find the x-values of all points where the function has any relative extrema. find the value(s) of any…

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

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

Answer

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).