what point of f(x)= e^(1-x^2) is concave down? I think it is -3/2
f' = -2x e^(1-x^2) zero when x = 0
f" = -2x (-2x e^(1-x^2)) -2 e^(1-x^2)
= (4 x^2 -2)e^(1-x^2)
when is that negative ?
e^whatever always +
so
when 2 (2x^2-1) is negative
x^2 always +
when x^2 <1/2
-1/sqrt 2 < x <1/sqrt 2 ?