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

the function ( f ) has a first derivative given by ( f^{prime}(x)=x(x - 3)^{2}(x + 1) ). at what values of ( x ) does ( f ) have a relative maximum? a -1 only b 0 only c -1 and 0 only d -1 and 3 only e -1, 0, and 3

the function ( f ) has a first derivative given by ( f^{prime}(x)=x(x - 3)^{2}(x + 1) ). at what values of ( x ) does ( f ) have a relative maximum? a -1 only b 0 only c -1 and 0 only d -1 and 3 only e -1, 0, and 3

Answer

Explanation:

Step1: Find critical points

Set (f^{\prime}(x)=0), so (x(x - 3)^{2}(x + 1)=0). Solving gives (x=-1), (x = 0), (x=3).

Step2: Use the first - derivative test

  • For (x<-1), let (x=-2), (f^{\prime}(-2)=(-2)(-2 - 3)^{2}(-2 + 1)=(-2)\times25\times(-1)=50>0).
  • For (-1<x<0), let (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), let (x = 1), (f^{\prime}(1)=(1)(1 - 3)^{2}(1 + 1)=(1)\times4\times2 = 8>0).
  • For (x>3), let (x = 4), (f^{\prime}(4)=(4)(4 - 3)^{2}(4 + 1)=(4)\times1\times5=20>0).

Since (f^{\prime}(x)) changes sign from positive to negative at (x=-1), (x=-1) is a relative maximum. At (x = 0), (f^{\prime}(x)) changes sign from negative to positive (a relative minimum). At (x = 3), (f^{\prime}(x)) does not change sign (no relative extremum as the multiplicity of the root (x = 3) is (2)).

Answer:

A. -1 only