A) You need to solve the equation f'(x) = 0 to find if the function has critical points such that:

`f'(x) = e^(-0.5x^2)*(-0.5x^2)'`

`f'(x) = e^(-0.5x^2)*(-x)`

`f'(x) = -x/sqrt(e^(x^2))`

You need to solve the equation `f'(x) = 0 =gt -x = 0 =gt x = 0.`

**Hence, the function has a critical value at x = 0.**

B) **Notice that the derivative f'(x)>0 for `x in (-oo,0),` hence the function increases over interval `(-oo,0).` **

C) **Notice that the derivative`f'(x)lt0` for `x in (0,oo), ` hence the function decreases over interval `(0,oo).` **

D) **The function reaches its maximum at x = 0.**

E) **The function has no minimum value.**

F) You need to solve f''(x) = 0 to verify if the function has inflection points such that:

`f''(x) = (-sqrt(e^(x^2)) + (2x^2*e^(x^2))/(2sqrt(e^(x^2))))/(e^(x^2))`

`f''(x) = (-sqrt(e^(x^2)) + x^2*e^(x^2)sqrt(e^(x^2)))/(e^(x^2))`

`f''(x) = (sqrt(e^(x^2))(-1 + x^2*e^(x^2)))/(e^(x^2))`

`f''(x) = 0 =gt -1 + x^2*e^(x^2) = ` 0

`x^2*e^(x^2) = 1 =gt e^(x^2) = 1/(x^2)`

`x^2 = ln(1/x^2)`

This is a transcendental equation and you may use graphical method to solve it:

**Notice that the intersections between the black curve and the red curve represents the inflection points of the graph of function, hence the x coordinates of inflection points are in interval (-1,1).**

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