What is the extreme f(x)=(x^2)/(2x-1) ?

The extreme f(x)=(x^2)/(2x-1) is Maximum(0,0),Minimum(1,1)

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extreme f(x)=(x^2)/(2x-1)

extreme points

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

Maximum (0, 0), Minimum (1, 1)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

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

Decreasing

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

Increasing

: 1 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{2 x - 1} : 0

Maximum (0, 0)

Plug

x = 1

into

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

Minimum (1, 1)

Maximum (0, 0), Minimum (1, 1)

ln (x - 4)

f (x) = (x - 5)^{2} - 9

f (x) = \frac{x - 2}{x - 9}

f (x) = - 9 x + 9

sin (π t)

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