How to find local min and max

How do you find the local min and max and the inflection point of the function y=xe^(-31x^2)

If you are in calculus you would take the derivative and then set it equal to zero.
use the product and chain rules respective you should get
yprime = e^(-31x^2) -61x2e^(-31x^2)
0 = e^(-31x^2) - 61x^2e(-31x^2)
factor
0 = (e^(-31x^2))(1-62x^2)
the first part is never zero
0 = 1 - 62x^2
62x^2 =1
x = + or minus √62/62
now graph it you will see that the positive one is a local max.

y = f(x) = xe^(-31x^2)
f'(x) = e^(-31x^2) - 62(x^2)e^(-31x^2)
f''(x) = -62xe^(-31x^2) - 124xe^(-31x^2) + 3844(x^3)e^(-31x^2) = (2xe^(-31x^2))(1922x^2 - 93)

Extrema occur when [a] f'(x) = 0 or [b] f'(x) does not exist.
e^(-31x^2) - 62(x^2)e^(-31x^2) = 0
1 - 62x^2 = 0
62x^2 = 1
x^2 = 1/62
x = ±√(1/62)
f(-1) = -e^(-31)
f(-√(1/62)) = -√(1/62)e^(-1/2) = -1/√(62e)
f(0) = 0
f(√(1/62)) = √(1/62)e^(-1/2) = 1/√(62e)
f(1) = e^(-31)
So (-√(1/62),-1/√(62e)) is a minimum and (√(1/62),1/√(62e)) is a maximum.

Inflection points occur when [1] f''(x) = 0 and [2] for some arbitrarily small c > 0 [i] f''(x ± c) ≠ 0 and [ii] f''(x - c) < 0 ⇔ f''(x + c) > 0. [2] Can be summarized to say that f''(x) changes sign at x.
(2xe^(-31x^2))(1922x^2 - 93) = 0
If 2xe^(-31x^2) = 0, then x = 0.
If 1922x^2 - 93 = 0, 1922x^2 = 93, x^2 = 93/1922, x = ±√(93/1922).
f''(-√(94/1922)) = (-2√(94/1922)e^(-31•94/1922))(94 - 93) < 0
f''(-√(92/1922)) = (-2√(92/1922)e^(-31•92/1922))(92 - 93) > 0
f''(√(92/1922)) = (2√(92/1922)e^(-31•92/1922))(92 - 93) < 0
f''(√(94/1922)) = (2√(92/1922)e^(-31•92/1922))(94 - 93) > 0
So, indeed, inflection points occur when x ∈ {-√(93/1922),0,√(93/1922)}.
These points are:
(-√(93/1922),-√(93/1922)e^(-2833/1922)…
(0,0)
(√(93/1922),√(93/1922)e^(-2833/1922))

Local min and max will be where the first derivative is -zero-.
Inflection point will be where the second derivative is zero.