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

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

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

extreme points

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

Maximum (- 1, - 1), Minimum (2, 2)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

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

Decreasing

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

Increasing

: 2 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, - 1)

Plug

x = 2

into

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

Minimum (2, 2)

Maximum (- 1, - 1), Minimum (2, 2)

f (x, y) = ∣ x y ∣

f (x) = - 2 cos (x) - 3 x

\frac{6 x}{x ^{2} + 1}

f (x) = - 1 - x^{2}, 0 \leq x \leq 1

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

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