What is the extreme f(x)= 1/(sqrt(2pi))e^{(-x^2)/2} ?

The extreme f(x)= 1/(sqrt(2pi))e^{(-x^2)/2} is Maximum(0, 1/(2^{1/2)pi^{1/2}})

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extreme f(x)= 1/(sqrt(2pi))e^{(-x^2)/2}

extreme points

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

Maximum (0, \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = 0

into

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

Maximum (0, \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}})

f (x) = \frac{2 x}{x ^{2} + 16}, - 5 \leq x \leq 5

f (x) = \frac{3 + x}{7 - x}, - 7 \leq x \leq 0

f (x) = e^{6 x^{2} + 4 y^{2} + 10}

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

f (x) = x (x + 3) e^{- 2 x}

What is the extreme f(x)= 1/(sqrt(2pi))e^{(-x^2)/2} ?
The extreme f(x)= 1/(sqrt(2pi))e^{(-x^2)/2} is Maximum(0, 1/(2^{1/2)pi^{1/2}})