For what values of x is #f(x)=x^3/e^x# concave or convex?

1 Answer
Jan 24, 2016

Concave up for all x>0 and concave down for all x<0


The concavity of a function is generally determined in a interval and not at a point. It is necessary to find the critical points at which the function changes its concavity. Accordingly , first find #f'(x) = (3x^2 e^x - x^3 e^x )/e^(2x) #


Now get the second derivative f" (x)= #((6x-3x^2)e^x - (3x^2 -x^3)e^x)/e^(2x)#

= #(x(x^2 -6x +6))/e^x#

Since f"(x) gives the slope of the graph of f'(x), hence when f" (x)>0, it would mean that f'(x) is increasing which would imply that f(x) is concave up at that point. Like wise when f" (x) <0, it would imply f'(x) is decreasing, indicating that f(x) would be concave down.

It would be seen from the f"(x) worked out above that for all negative values of x, f"(x)<0 , meaning thereby that f(x) would be concave down for all negative x.

Also for all positive values of x, f"(x) >0, meaning thereby that f(x) would be concave up for all positive x.