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

The extreme f(x)=e^{x^2-x} is Minimum(1/2 , 1/(\sqrt[4]{e)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = \frac{1}{2}

into

f (x) = x^{3} - 3 x^{2} + 7 x - 10

f (x) = - 10 x^{2} - 40 x - 2

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

10 x^{2} - \frac{250}{x}

f (x) = x^{7} (x + 5)^{8}, - 10 \leq x \leq 10

What is the extreme f(x)=e^{x^2-x} ?
The extreme f(x)=e^{x^2-x} is Minimum(1/2 , 1/(\sqrt[4]{e)})