if ( f(x)=x(x + 1)(x - 2)^2 ), then the graph of ( f ) has inflection points when ( x= ) \n\na -1 only \nb 2…

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