What is the extreme y=xe^{-2x^2} ?

The extreme y=xe^{-2x^2} is Minimum(-1/2 ,-1/(2sqrt(e))),Maximum(1/2 , 1/(2sqrt(e)))

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extreme y=xe^{-2x^2}

extreme points

y = x e^{- 2 x^{2}}

Minimum (- \frac{1}{2}, - \frac{1}{2 e}), Maximum (\frac{1}{2}, \frac{1}{2 e})

Solution steps

f^{'} (x) = - e^{- 2 x^{2}} (2 x + 1) (2 x - 1)

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{2},

Increasing

: - \frac{1}{2} < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < \infty

Plug

x = - \frac{1}{2}

into

x e^{- 2 x^{2}} : - \frac{1}{2 e}

Minimum (- \frac{1}{2}, - \frac{1}{2 e})

Plug

x = \frac{1}{2}

into

x e^{- 2 x^{2}} : \frac{1}{2 e}

Maximum (\frac{1}{2}, \frac{1}{2 e})

Minimum (- \frac{1}{2}, - \frac{1}{2 e}), Maximum (\frac{1}{2}, \frac{1}{2 e})

f (x) = \frac{1 + 3 x ^{2} - x ^{3}}{x ^{2}}

f (x) = \frac{( - 4 x ^{2} + 100 )}{5 x - 25}

\frac{x ^{2}}{x - 2}

4 x + 2 y = 10

\frac{( 4 x - 1 )}{2 x + 9}

What is the extreme y=xe^{-2x^2} ?
The extreme y=xe^{-2x^2} is Minimum(-1/2 ,-1/(2sqrt(e))),Maximum(1/2 , 1/(2sqrt(e)))