Question

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 relative maximum?

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
• For \(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:

\(x=-1\)