Question

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

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
• For \(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