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

Jan 24, 2016

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

#### Explanation:

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 $f ' \left(x\right) = \frac{3 {x}^{2} {e}^{x} - {x}^{3} {e}^{x}}{e} ^ \left(2 x\right)$

=$\frac{{x}^{2} \left(3 - x\right)}{e} ^ x$

Now get the second derivative f" (x)= $\frac{\left(6 x - 3 {x}^{2}\right) {e}^{x} - \left(3 {x}^{2} - {x}^{3}\right) {e}^{x}}{e} ^ \left(2 x\right)$

= $\frac{x \left({x}^{2} - 6 x + 6\right)}{e} ^ x$

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.