Question

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

Answer 1

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)`.