Is #f(x) =e^x-x^3-3x# concave or convex at #x=0#?

1 Answer
Aug 23, 2017

Concave

Explanation:

Compute the second derivative.

#f'(x) = e^x - 3x^2 - 3#

#f''(x) = e^x - 6x#

At #x = 0#:

#f''(0) = e^0 - 6(0) = 1#

Since this value is positive, the function is concave at #x = 0#.

Hopefully this helps!