How do you find the inflection point, concave up and down for f(x)=x^3-3x^2+3?

1 Answer
Bill K.
Jun 3, 2015

The first derivative is f'(x)=3x^2-6x and the second derivative is f''(x)=6x-6=6(x-1). The second derivative is negative when x<1, positive when x>1, and zero when x=1 (and of course changes sign as x increases "through" x=1).

That means the graph of f is concave down when x<1, concave up when x>1, and has an inflection point at x=1. The coordinates of the inflection point are (x,y)=(1,f(1))=(1,1).

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