Question

if ( f(x)=x(x + 1)(x - 2)^2 ), then the graph of ( f ) has inflection points when ( x= )

a -1 only
b 2 only
c -1 and 0 only
d -1 and 2 only
e -1, 0, and 2 only

Explanation:

Step1: Find the roots of \(f''(x)\)

The roots of \(f''(x)=x(x + 1)(x - 2)^2\) are \(x=-1\), \(x = 0\), and \(x=2\) (since when \(x=-1\), \(f''(-1)=(-1)(-1 + 1)(-1 - 2)^2=0\); when \(x = 0\), \(f''(0)=(0)(0 + 1)(0 - 2)^2=0\); when \(x=2\), \(f''(2)=(2)(2 + 1)(2 - 2)^2=0\)).

Step2: Test the sign - change of \(f''(x)\)

• For \(x=-1\):

Choose a test - point \(x=-2\) (to the left of \(x = - 1\)): \(f''(-2)=(-2)(-2 + 1)(-2 - 2)^2=(-2)(-1)(16)=32>0\).
Choose a test - point \(x=-\frac{1}{2}\) (to the right of \(x=-1\)): \(f''(-\frac{1}{2})=(-\frac{1}{2})(-\frac{1}{2}+1)(-\frac{1}{2}-2)^2=(-\frac{1}{2})(\frac{1}{2})(\frac{25}{4})=-\frac{25}{16}<0\). There is a sign - change at \(x=-1\).

• For \(x = 0\):

Choose a test - point \(x=-\frac{1}{2}\) (to the left of \(x = 0\)): \(f''(-\frac{1}{2})=(-\frac{1}{2})(-\frac{1}{2}+1)(-\frac{1}{2}-2)^2=-\frac{25}{16}<0\).
Choose a test - point \(x = 1\) (to the right of \(x = 0\)): \(f''(1)=(1)(1 + 1)(1 - 2)^2=(1)(2)(1)=2>0\). There is a sign - change at \(x = 0\).

• For \(x=2\):

Choose a test - point \(x = 1\) (to the left of \(x = 2\)): \(f''(1)=(1)(1 + 1)(1 - 2)^2=2>0\).
Choose a test - point \(x=3\) (to the right of \(x = 2\)): \(f''(3)=(3)(3 + 1)(3 - 2)^2=(3)(4)(1)=12>0\). There is no sign - change at \(x = 2\).

Since the graph of \(y = f(x)\) has an inflection point when \(f''(x)\) changes sign, and \(f''(x)\) changes sign at \(x=-1\) and \(x = 0\) but not at \(x = 2\).

Answer:

C. -1 and 0 only