What is the extreme f(x)=(5x)/(e^{2x^2+3)} ?

The extreme f(x)=(5x)/(e^{2x^2+3)} is Minimum(-1/2 ,-5/(2e^{7/2)}),Maximum(1/2 , 5/(2e^{7/2)})

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extreme f(x)=(5x)/(e^{2x^2+3)}

extreme points

f (x) = \frac{5 x}{e ^{2 x^{2} + 3}}

Minimum (- \frac{1}{2}, - \frac{5}{2 e ^{\frac{7}{2}}}), Maximum (\frac{1}{2}, \frac{5}{2 e ^{\frac{7}{2}}})

Solution steps

f^{'} (x) = 5 e^{- 2 x^{2} - 3} (1 - 4 x^{2})

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

\frac{5 x}{e ^{2 x^{2} + 3}} : - \frac{5}{2 e ^{\frac{7}{2}}}

Minimum (- \frac{1}{2}, - \frac{5}{2 e ^{\frac{7}{2}}})

Plug

x = \frac{1}{2}

into

\frac{5 x}{e ^{2 x^{2} + 3}} : \frac{5}{2 e ^{\frac{7}{2}}}

Maximum (\frac{1}{2}, \frac{5}{2 e ^{\frac{7}{2}}})

Minimum (- \frac{1}{2}, - \frac{5}{2 e ^{\frac{7}{2}}}), Maximum (\frac{1}{2}, \frac{5}{2 e ^{\frac{7}{2}}})

+ x^{2} + 4 x + 6

f (x) = x^{\frac{1}{3}} + b x^{\frac{4}{3}}

(1 + x y) (x + y)

y = \frac{2 x}{x ^{2} + 1}

f (x, y) = 6 - 2 x - 3 y

What is the extreme f(x)=(5x)/(e^{2x^2+3)} ?
The extreme f(x)=(5x)/(e^{2x^2+3)} is Minimum(-1/2 ,-5/(2e^{7/2)}),Maximum(1/2 , 5/(2e^{7/2)})