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

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)

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)