What is the extreme f(x)=e^{x^2-3x-1} ?

The extreme f(x)=e^{x^2-3x-1} is Minimum(3/2 , 1/(e^{13/4)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=e^{x^2-3x-1}

extreme points

f (x) = e^{x^{2} - 3 x - 1}

Minimum (\frac{3}{2}, \frac{1}{e ^{\frac{13}{4}}})

Solution steps

f^{'} (x) = e^{x^{2} - 3 x - 1} (2 x - 3)

Find intervals:

Decreasing

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

Increasing

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

Plug

x = \frac{3}{2}

into

e^{x^{2} - 3 x - 1} : \frac{1}{e ^{\frac{13}{4}}}

Minimum (\frac{3}{2}, \frac{1}{e ^{\frac{13}{4}}})

What is the extreme f(x)=e^{x^2-3x-1} ?
The extreme f(x)=e^{x^2-3x-1} is Minimum(3/2 , 1/(e^{13/4)})