the function f has a first derivative given by f(x)=x(x - 3)^2(x + 1). at what values of x does f have a…

Explanation:

Step1: Find critical points

Set (f^{\prime}(x)=x(x - 3)^{2}(x + 1)=0). Using the zero - product property (a\times b\times c\times d = 0) implies (a = 0) or (b = 0) or (c = 0) or (d = 0). So (x=0), (x = 3), (x=-1) are the critical points.

Step2: Use the first - derivative test

Create a sign chart for (f^{\prime}(x)).

• Choose test points: for (x<-1) (say (x=-2)), (f^{\prime}(-2)=(-2)(-2 - 3)^{2}(-2 + 1)=(-2)\times25\times(-1)=50>0).
• For (-1<x<0) (say (x =-\frac{1}{2})), (f^{\prime}(-\frac{1}{2})=(-\frac{1}{2})(-\frac{1}{2}-3)^{2}(-\frac{1}{2}+1)=(-\frac{1}{2})\times(\frac{-7}{2})^{2}\times\frac{1}{2}=(-\frac{1}{2})\times\frac{49}{4}\times\frac{1}{2}=-\frac{49}{16}<0).
• For (0<x<3) (say (x = 1)), (f^{\prime}(1)=(1)(1 - 3)^{2}(1 + 1)=(1)\times4\times2 = 8>0).
• For (x>3) (say (x = 4)), (f^{\prime}(4)=(4)(4 - 3)^{2}(4 + 1)=(4)\times1\times5=20>0).

Since the function (f(x)) changes from increasing ((f^{\prime}(x)>0)) to decreasing ((f^{\prime}(x)<0)) at (x=-1), and changes from decreasing ((f^{\prime}(x)<0)) to increasing ((f^{\prime}(x)>0)) at (x = 0), and the sign of (f^{\prime}(x)) does not change at (x = 3) (because the factor ((x - 3)^{2}\geqslant0) and when (x) passes through (3), the sign of (f^{\prime}(x)) is non - negative on both sides of (x = 3)).

Answer:

(x=-1)