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

The extreme f(x)=(|x|)/(2-x) is Minimum(0,0)

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extreme f(x)=(|x|)/(2-x)

extreme points

f (x) = \frac{∣ x ∣}{2 - x}

Minimum (0, 0)

Solution steps

Find the critical points:

x = 0

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = 0

into

\frac{∣ x ∣}{2 - x} \Rightarrow y = 0

Minimum (0, 0)

f (x, y) = 4 x y e^{y - x}

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

f (x) = 3 x + 3 y

f (x, y) = 2 x^{3} y^{2} + \frac{5}{x} - \frac{1}{y} + 4

f (x) = \frac{- 7}{x ^{2} - 4}

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