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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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)

ln (\frac{x}{3})

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

f (4) = \frac{24}{x ( x + 2 )}

f (x) = 3 x^{2} + 6 x + 2

\frac{x}{x ^{2} + 3}

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