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

The extreme f(x)=(5e^x+5e^{-x})/2 is Minimum(0,5)

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

extreme points

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

Minimum (0, 5)

Solution steps

f^{'} (x) = \frac{1}{2} (5 e^{x} - 5 e^{- x})

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

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

Minimum (0, 5)

What is the extreme f(x)=(5e^x+5e^{-x})/2 ?
The extreme f(x)=(5e^x+5e^{-x})/2 is Minimum(0,5)