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

Solution

extreme points

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

Solution

Minimum (0, 1)

Solution steps

f^{'} (x) = 4 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

2 x^{2} + 1 : 1

Minimum (0, 1)

Popular Examples

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f (x) = \frac{( 4 x ^{2} )}{x - 2}

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

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

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

extreme x^{1/9}+1 extreme points

x^{\frac{1}{9}} + 1

extreme f(x)=(3x)/(x-2)extreme points

f (x) = \frac{3 x}{x - 2}

Frequently Asked Questions (FAQ)

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