What is the f(x)=\sqrt[3]{(x^2+1)/(x^2-1)} ?

The f(x)=\sqrt[3]{(x^2+1)/(x^2-1)} is

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f(x)=\sqrt[3]{(x^2+1)/(x^2-1)}

Domain : x < - 1 or - 1 < x < 1 or x > 1

Range : f (x) \leq - 1 or f (x) > 1

Y Intercepts : (0, - 1)

Asymptotes : Vertical x = - 1, x = 1, Horizontal y = 1

Extreme Points : Maximum (0, - 1)

Interval Notation

Domain : (- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)

Range : (- \infty, - 1] \cup (1, \infty)

Solution steps

Domain of

Range of

Axis interception points of

Y Intercepts

: (0, - 1)

Asymptotes of

Vertical

: x = - 1, x = 1,

Horizontal

: y = 1

Extreme Points of

Maximum

(0, - 1)

y = \frac{1}{3 x - 2}

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

f (x) = - x^{2} + 7 x - 6

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

f (x) = 3 x - 11

What is the f(x)=\sqrt[3]{(x^2+1)/(x^2-1)} ?
The f(x)=\sqrt[3]{(x^2+1)/(x^2-1)} is