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Popular >

extreme 2x^2ln(x)

Solution

extreme points

2 x^{2} ln (x)

Solution

Minimum (\frac{1}{e}, - \frac{1}{e})

Solution steps

f^{'} (x) = 2 (2 x ln (x) + x)

Find intervals:

Decreasing

: 0 < x < \frac{1}{e},

Increasing

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

Plug

x = \frac{1}{e}

into

2 x^{2} ln (x) : - \frac{1}{e}

Minimum (\frac{1}{e}, - \frac{1}{e})

Popular Examples

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extreme f(x,y)=7+x+4y-x^2-4y^2 extreme points

f (x, y) = 7 + x + 4 y - x^{2} - 4 y^{2}

extreme f(x)=e^{x^2-4x-1},-4<= x<= 4 extreme points

f (x) = e^{x^{2} - 4 x - 1}, - 4 \leq x \leq 4

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

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

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f (x) = (x - 2) (x + 1) (x + 5)

Frequently Asked Questions (FAQ)

What is the extreme 2x^2ln(x) ?
The extreme 2x^2ln(x) is Minimum(1/(sqrt(e)),-1/e)