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

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

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extreme (|1-x^2|)/x

extreme points

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

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

Solution steps

Find the critical points:

x = - 1, x = 1

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = - 1

into

\frac{1 - x ^{2}}{x} \Rightarrow y = 0

Maximum (- 1, 0)

Plug the extreme point

x = 1

into

\frac{1 - x ^{2}}{x} \Rightarrow y = 0

Minimum (1, 0)

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

f (x) = 3 x^{2} - 18 x + 24

f (x, y) = x^{3} + y^{3} - 24 x y

f (x, y) = - 2 x^{2} - 2 y^{2} - 4 x y + 5 x + 4 y

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

(x^{2} + 12) (4 - x^{2})

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