What is the extreme f(χ)=χ^2e^{-4χ^2} ?

The extreme f(χ)=χ^2e^{-4χ^2} is Maximum(-1/2 , 1/(4e)),Minimum(0,0),Maximum(1/2 , 1/(4e))

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extreme f(χ)=χ^2e^{-4χ^2}

extreme points

f (χ) = χ^{2} e^{- 4 χ^{2}}

Maximum (- \frac{1}{2}, \frac{1}{4 e}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{4 e})

Solution steps

f^{'} (χ) = - 2 e^{- 4 χ^{2}} χ (2 χ + 1) (2 χ - 1)

Find intervals:

Increasing

: - \infty < χ < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < χ < 0,

Increasing

: 0 < χ < \frac{1}{2},

Decreasing

: \frac{1}{2} < χ < \infty

Plug

χ = - \frac{1}{2}

into

χ^{2} e^{- 4 χ^{2}} : \frac{1}{4 e}

Maximum (- \frac{1}{2}, \frac{1}{4 e})

Plug

χ = 0

into

χ^{2} e^{- 4 χ^{2}} : 0

Minimum (0, 0)

Plug

χ = \frac{1}{2}

into

χ^{2} e^{- 4 χ^{2}} : \frac{1}{4 e}

Maximum (\frac{1}{2}, \frac{1}{4 e})

Maximum (- \frac{1}{2}, \frac{1}{4 e}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{4 e})

f (x) = \frac{y - 800}{- 2}

(x - 1) e^{\frac{1}{x}}

f (x) = x^{4} + x^{2} (y - 2) + 3 (y - 1)^{2}

f (x, y) = 2 e^{x} ln (1 + y)

f (x) = 9 (x - 4)^{2} + 5

What is the extreme f(χ)=χ^2e^{-4χ^2} ?
The extreme f(χ)=χ^2e^{-4χ^2} is Maximum(-1/2 , 1/(4e)),Minimum(0,0),Maximum(1/2 , 1/(4e))