What is the extreme y=(sqrt(1-x^2))/x ?

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

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extreme y=(sqrt(1-x^2))/x

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1

Plug

x = - 1

into

\frac{1 - x ^{2}}{x} : 0

Maximum (- 1, 0)

Plug

x = 1

into

\frac{1 - x ^{2}}{x} : 0

Minimum (1, 0)

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

x^{4} - 5 x^{2} - 4

F (X, Y) = 2 X + 3 Y X^{2}

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

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

f (x) = - 2 x^{2} - 1

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