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

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

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

extreme points

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

Maximum (0, 1), Minimum (1, 2)

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}{2 x - 1} : 1

Maximum (0, 1)

Plug

x = 1

into

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

Minimum (1, 2)

Maximum (0, 1), Minimum (1, 2)

\frac{1}{2} sin (x + \frac{π}{4})

f (x) = x^{\frac{6}{7}} - 7

f (x) = x^{3} + y^{3} - 3 x^{2} - 3 y^{2} - 9 x

f (x) = \frac{( x ^{3} )}{3} - 2 x^{2} - 5 x

f (x, y) = - 4 x^{2} + y (8 - y)

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