Question 666176
y'(x) = -2x e^(-x^2)

Your critical point is where 0 = -2xe^(-x^2)

Since e will never be 0, we have to rely on -2x, which of course, 0 makes it 0.

So <------------0---------------> If we plug in a negative number we get a positive, if we plug in a positive we get a negative so we get:

<---negative----0------positive---->

From -oo to 0 decreasing,  trough at 0,  increasing to infinity. No absolute max, since we have a trough. A peak would give us an abs max.


y''(x) =  -2x*-2x*e^(-x^2) +  e^(-x^2)*-2 (by product rule)

=4x^2e^(-x^2) - 2e^(-x^2)

= (4x^2-1)(e^(-x^2))

4x^2 -1 = 0

4x^2 = 1

x^2 = 1/4

x = +- 1/2

<---- -1/2 ------------ 1/2 ---------->

choose -1 for the left side.

We get positive.

choose 0, we get negative

choose 1 we get positive


<--positive -- -1/2  ---- negative ---- 1/2 ---- positive---->


Hence we have concave up from negative infinity to -1/2

concave down from -1/2 to 1/2

concave up from 1/2 to oo

points of inflection are x= -1/2 and 1/2