en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme 2^{x^2-5}

Solution

extreme points

2^{x^{2} - 5}

Solution

Minimum (0, \frac{1}{32})

Solution steps

f^{'} (x) = ln (2) \cdot 2^{x^{2} - 4} x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

2^{x^{2} - 5} : \frac{1}{32}

Minimum (0, \frac{1}{32})

Popular Examples

extreme f(x)=x^{3/4}-4x^{1/4}extreme points

f (x) = x^{\frac{3}{4}} - 4 x^{\frac{1}{4}}

x = \frac{2}{x}

extreme f(x)=sqrt(420-x)x extreme points

f (x) = 420 - x x

extreme f(x)=3x^2-10x-25 extreme points

f (x) = 3 x^{2} - 10 x - 25

extreme S(x,y)=3x-2y+(3x-5y)-2x+[3y-(2x-7y)+4x]extreme points

S (x, y) = 3 x - 2 y + (3 x - 5 y) - 2 x + [3 y - (2 x - 7 y) + 4 x]

Frequently Asked Questions (FAQ)

What is the extreme 2^{x^2-5} ?
The extreme 2^{x^2-5} is Minimum(0, 1/32)