What is the extreme f(x)=((-2x^2+5x-1))/(2x-1) ?

The extreme f(x)=((-2x^2+5x-1))/(2x-1) is None

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extreme f(x)=((-2x^2+5x-1))/(2x-1)

extreme points

f (x) = \frac{( - 2 x ^{2} + 5 x - 1 )}{2 x - 1}

None

Solution steps

f^{'} (x) = \frac{- 4 x ^{2} + 4 x - 3}{( 2 x - 1 ) ^{2}}

Find intervals:

None

None

f (x, y) = \frac{1}{3} x^{2} + 2 x y + 3 y^{2} + 4 x - 1

f (x, y) = 4 x + 6 y - x^{2} - y^{2} + 8

f (x) = x^{3} + 6 x^{2} + 8

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

f (x) = \frac{4 x - 12}{( x - 2 ) ^{2}}

What is the extreme f(x)=((-2x^2+5x-1))/(2x-1) ?
The extreme f(x)=((-2x^2+5x-1))/(2x-1) is None