What is the extreme ((2x+1))/((e^{x^2))} ?

The extreme ((2x+1))/((e^{x^2))} is Minimum(-1,-1/e),Maximum(1/2 , 2/(e^{1/4)})

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extreme ((2x+1))/((e^{x^2))}

extreme points

\frac{( 2 x + 1 )}{( e ^{x^{2}} )}

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

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

Decreasing

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

Plug

x = - 1

into

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

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

Plug

x = \frac{1}{2}

into

\frac{2 x + 1}{e ^{x^{2}}} : \frac{2}{e ^{\frac{1}{4}}}

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

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

f (y) = 3 y^{2} - 3 x^{2} - 6 y

f (x, y) = x^{3} y + 12 y x^{2} - 8 y

f (x, y) = 2 e^{- y} (x^{2} + y^{2}) + 5

f (x) = \frac{5 - x ^{2}}{x}

x^{2} - 14 x + 2

What is the extreme ((2x+1))/((e^{x^2))} ?
The extreme ((2x+1))/((e^{x^2))} is Minimum(-1,-1/e),Maximum(1/2 , 2/(e^{1/4)})