What is the extreme p(x)=e^{x-3}+e^{-x} ?

The extreme p(x)=e^{x-3}+e^{-x} is Minimum(3/2 , 2/(e^{3/2)})

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extreme p(x)=e^{x-3}+e^{-x}

extreme points

p (x) = e^{x - 3} + e^{- x}

Minimum (\frac{3}{2}, \frac{2}{e ^{\frac{3}{2}}})

Solution steps

f^{'} (x) = e^{x - 3} - e^{- x}

Find intervals:

Decreasing

: - \infty < x < \frac{3}{2},

Increasing

: \frac{3}{2} < x < \infty

Plug

x = \frac{3}{2}

into

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

Minimum (\frac{3}{2}, \frac{2}{e ^{\frac{3}{2}}})

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

f (x, y) = 4 x^{2} + 8 y^{2}

f (x) = \frac{2 x + 1 - 18}{x}

f (x, y) = - 11 \cdot y^{2} + (x + 16)^{2} + 1

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

What is the extreme p(x)=e^{x-3}+e^{-x} ?
The extreme p(x)=e^{x-3}+e^{-x} is Minimum(3/2 , 2/(e^{3/2)})