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Finding a Potential Point of Inflection

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The second derivative of the function f(x)=x3+3x2−6x+1f(x)= x^3 + 3x^2 - 6x + 1 is f′′(x)=6x+6f''(x)= 6x+6. For what value of xx is f′′(x)=0f''(x)=0?

A

x=1x=1

B

x=0x=0

C

x=−6x=-6

D

x=−1x=-1

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