en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(2x^2+1)^2

Solution

extreme points

f (x) = (2 x^{2} + 1)^{2}

Solution

Minimum (0, 1)

Solution steps

f^{'} (x) = 8 x (2 x^{2} + 1)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

(2 x^{2} + 1)^{2} : 1

Minimum (0, 1)

Popular Examples

extreme f(x,y)=(100)/(1+x^2+y^2)extreme points

f (x, y) = \frac{100}{1 + x ^{2} + y ^{2}}

extreme f(x)= 1/6 x^3-x^2+6x+11 extreme points

f (x) = \frac{1}{6} x^{3} - x^{2} + 6 x + 11

extreme f(x,y)=x^2+y^2+8x-2y extreme points

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

extreme f(x)=230+8x^3+x^4 extreme points

f (x) = 230 + 8 x^{3} + x^{4}

extreme f(x,y)=e^{-(x^2+y^2)}(x^2+2y^2)extreme points

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

Frequently Asked Questions (FAQ)

What is the extreme f(x)=(2x^2+1)^2 ?
The extreme f(x)=(2x^2+1)^2 is Minimum(0,1)