Given f(x)=2-(x^3/6)-x^2. On what interval(s) is the graph of f concave upwards?
The derivative is (-x^2/2)-2x but do I set that equal to 0?

Question

Given f(x)=2-(x^3/6)-x^2. On what interval(s) is the graph of f concave upwards? 
The derivative is (-x^2/2)-2x but do I set that equal to 0?

anonymous · Answer

Okay well to find concavity you have to work with the second derivative of the function, so I would take the derivative of (-x^2/2)-2x and then set that equal to 0. Then the values that you get you would put in a sign analysis. Concave up is when the second derivative is increasing.

anonymous · Answer

Now I remember. Thank you.