What is the extreme f(x)=ln(x)-2x^2 ?

The extreme f(x)=ln(x)-2x^2 is Maximum(1/2 ,-ln(2)-1/2)

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extreme f(x)=ln(x)-2x^2

extreme points

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

Maximum (\frac{1}{2}, - ln (2) - \frac{1}{2})

Solution steps

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

Find intervals:

Increasing

: 0 < x < \frac{1}{2},

Decreasing

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

Plug

x = \frac{1}{2}

into

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

Maximum (\frac{1}{2}, - ln (2) - \frac{1}{2})

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

f (x, y) = - 2 x^{3} - 5 x^{2} + 4 x y - y^{2} + 4 y - 7

f (x, y) = 5 x^{2} + 5 y^{2} + 5 x y - 10 x - 5 y + 18

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

x^{2} + x y + y^{2} - 2 x - y

What is the extreme f(x)=ln(x)-2x^2 ?
The extreme f(x)=ln(x)-2x^2 is Maximum(1/2 ,-ln(2)-1/2)