What is the f(x)=sqrt((x(3+x)-4)/(x^2-1)) ?

The f(x)=sqrt((x(3+x)-4)/(x^2-1)) is

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f(x)=sqrt((x(3+x)-4)/(x^2-1))

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

Domain : x \leq - 4 or - 1 < x < 1 or x > 1

Range : 0 \leq f (x) < 1 or 1 < f (x) < \frac{5}{2} or f (x) > \frac{5}{2}

X Intercepts : (- 4, 0), Y Intercepts : (0, 2)

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

Extreme Points : None

Interval Notation

Domain : (- \infty, - 4] \cup (- 1, 1) \cup (1, \infty)

Range : [0, 1) \cup (1, \frac{5}{2}) \cup (\frac{5}{2}, \infty)

Solution steps

Domain of

\frac{x ( 3 + x ) - 4}{x ^{2} - 1} : x \leq - 4 or - 1 < x < 1 or x > 1

Range of

\frac{x ( 3 + x ) - 4}{x ^{2} - 1} : 0 \leq f (x) < 1 or 1 < f (x) < \frac{5}{2} or f (x) > \frac{5}{2}

Axis interception points of

\frac{x ( 3 + x ) - 4}{x ^{2} - 1} :

X Intercepts

: (- 4, 0),

Y Intercepts

: (0, 2)

Asymptotes of

\frac{x ( 3 + x ) - 4}{x ^{2} - 1} :

Vertical

: x = - 1,

Horizontal

: y = 1

Extreme Points of

\frac{x ( 3 + x ) - 4}{x ^{2} - 1} :

None

y = x^{2} - 64

36 y^{2} = (x^{2} - 4)^{3}, 2 \leq x \leq 3

f (x) = - x^{2} - 6 x - 9

y = \frac{3}{4} x^{2}

f (x) = 4 x^{2} - 4 x + 1

What is the f(x)=sqrt((x(3+x)-4)/(x^2-1)) ?
The f(x)=sqrt((x(3+x)-4)/(x^2-1)) is