Question 1100669: Find the open intervals where the function is concave up and concave down.
f(x)= x^3 -3x^2 -1
Answer: concave up (1,infinity)
concave down (-infinity, 1)
i dont understand because i got 0,2 for the critical points when you equal the whole derivative to zero. So please explain how to do it by hand without the calculator. Thanks.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! f(x)= x^3 -3x^2 -1
It's true that the first derivative shows critical points of 0, 2
The inflection point uses the second derivative, which is 6x-6
at 0, f''(x) is -6 and at (0.5) it is -3 and at 1 it is 0 and at 1.5 it is +3
Therefore, the concavity changes at 1. The critical points will be where there are local maxima or minima; the concavity changes are in between somewhere.
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