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

The extreme f(x)=x+e^{-3x} is Minimum((ln(3))/3 ,(ln(3)+1)/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Minimum (\frac{ln ( 3 )}{3}, \frac{ln ( 3 ) + 1}{3})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < \frac{ln ( 3 )}{3},

Increasing

: \frac{ln ( 3 )}{3} < x < \infty

Plug

x = \frac{ln ( 3 )}{3}

into

x + e^{- 3 x} : \frac{ln ( 3 ) + 1}{3}

Minimum (\frac{ln ( 3 )}{3}, \frac{ln ( 3 ) + 1}{3})

y = (6 x - 5) e^{6 - 2 x}

f (x, y) = 1 - x^{2} + 1 - y^{2}

- x e^{- \frac{x}{4}}

f (x) = x^{2} - 2 x, 1 \leq x \leq 3

f (x) = x^{2} - 2 x, 1 \leq x \leq 4

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