What is the extreme (x^4)/(x^2-1) ?

The extreme (x^4)/(x^2-1) is Minimum(-sqrt(2),4),Maximum(0,0),Minimum(sqrt(2),4)

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extreme (x^4)/(x^2-1)

extreme

Minimum (- 2, 4), Maximum (0, 0), Minimum (2, 4)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{x ^{4}}{x ^{2} - 1} : 4

Minimum (- 2, 4)

Plug

x = 0

into

\frac{x ^{4}}{x ^{2} - 1} : 0

Maximum (0, 0)

Plug

x = 2

into

\frac{x ^{4}}{x ^{2} - 1} : 4

Minimum (2, 4)

Minimum (- 2, 4), Maximum (0, 0), Minimum (2, 4)

What is the extreme (x^4)/(x^2-1) ?
The extreme (x^4)/(x^2-1) is Minimum(-sqrt(2),4),Maximum(0,0),Minimum(sqrt(2),4)