Question

What are the critical points and also inflection points of `3e^(-2x(^2))`?

Answer 1

Assuming that the function is: `f(x) = 3e^(-2x^2)`, the only critical number is: `0`, and the inflection points are: `(1/sqrt12, 3e^(-1/6))` and `(-1/sqrt12, 3e^(-1/6))`

Explanation:

`f(x) = 3e^(-2x^2)`

`f'(x) = 3e^(-2x^2) * (-4x) = -12xe^(-2x^2)`

`f'(x)` is never undefined and is `0` when `x=0`.

The only critical number is `0`.

`f''(x) = -12e^(-2x^2) -12x(-12xe^(-2x^2))`

`= -12e^(-2x^2)(1-12x^2)`

#f''(x) is never undefined, so the only chance it has to change signs is when it is zero.

Solving: `-12e^(-2x^2)(1-12x^2) = 0`, gives us

`x = +-1/sqrt12`

Observe that the sign of `f''(x)` is the opposite of the sign of `1-12x^2` which does, indeed, change sign at both `+1/sqrt12` and `-1/sqrt12`

`f` is an even function so we calculate: `f( +-1/sqrt12) = 3e^(-2(1/12)) = 3e^(-1/6)`

The inflection points are: `(1/sqrt12, 3e^(-1/6))` and `(-1/sqrt12, 3e^(-1/6))`

(Simplify and rationalize if you prefer.)