Find where the graph is increasing and concave up for f(x)=e^(-x^2)

Expert Answers
embizze eNotes educator| Certified Educator

Given `f(x)=e^(-x^2)` :

`f'(x)=-2xe^(-x^2)`  Since `e^(-x^2)>0` for all x, the derivative is positive on `(-oo,0)` and negative on `(0,oo)`

The function is increasing for x<0, has a global maximum at x=0, and is decreasing for x>0


Since `e^(-x^2)>0` for all x we consider `4x^2-2`

`4x^2-2=0==> x^2=1/2` or `x=+-1/sqrt(2)`

Checking test values we find `f''(x)>0 (-oo,-1/sqrt(2));f''(x)<0 (-1/sqrt(2),1/sqrt(2));f''(x)>0 (1/sqrt(2),oo)` while `f''(x)=0` at `x=+-1/sqrt(2)`

So the function is concave up on `(-oo,-1/sqrt(2))` , has an inflection point at `x=-1/sqrt(2)` , is concave down on `(-1/sqrt(2),1/sqrt(2))` , has an inflection point at `x=1/sqrt(2)` , and is concave up on `(1/sqrt(2),oo)`


The graph is increasing on `(-oo,0)` and concave up on `(-oo,-1/sqrt(2)),(1/sqrt(2),oo)`


The graph: