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

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

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

extreme points

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

Minimum (- 2, 2), Minimum (2, 2)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < - 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

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

Minimum (- 2, 2)

Plug

x = 2

into

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

Minimum (2, 2)

Minimum (- 2, 2), Minimum (2, 2)

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

f (x, y) = 2 x^{4} + y^{4} - 2 x^{2} - 2 y^{2} + 3

f (x) = 9 - 2 x + 4 y - x^{2} - 4 y^{2}

f (x) = x^{3} + 2 x^{2} + 1

\frac{5 x ^{2}}{x ^{2} + 16}

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