What is the f(x)=(4x^2-1)/(6x^3-x|2-3x|-1) ?

The f(x)=(4x^2-1)/(6x^3-x|2-3x|-1) is

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f(x)=(4x^2-1)/(6x^3-x|2-3x|-1)

f (x) = \frac{4 x ^{2} - 1}{6 x ^{3} - x ∣ 2 - 3 x ∣ - 1}

Domain : x < - \frac{1}{3} or - \frac{1}{3} < x < - \frac{1}{2} or - \frac{1}{2} < x < \frac{1}{3} or x > \frac{1}{3}

X Intercepts : (\frac{1}{2}, 0), Y Intercepts : (0, 1)

Asymptotes : Horizontal y = 0

Extreme Points : None

Interval Notation

Domain : (- \infty, - \frac{1}{3}) \cup (- \frac{1}{3}, - \frac{1}{2}) \cup (- \frac{1}{2}, \frac{1}{3}) \cup (\frac{1}{3}, \infty)

Solution steps

Domain of

\frac{4 x ^{2} - 1}{6 x ^{3} - x ∣ 2 - 3 x ∣ - 1} : x < - \frac{1}{3} or - \frac{1}{3} < x < - \frac{1}{2} or - \frac{1}{2} < x < \frac{1}{3} or x > \frac{1}{3}

Axis interception points of

\frac{4 x ^{2} - 1}{6 x ^{3} - x ∣ 2 - 3 x ∣ - 1} :

X Intercepts

: (\frac{1}{2}, 0),

Y Intercepts

: (0, 1)

Asymptotes of

\frac{4 x ^{2} - 1}{6 x ^{3} - x ∣ 2 - 3 x ∣ - 1} :

Horizontal

: y = 0

Extreme Points of

\frac{4 x ^{2} - 1}{6 x ^{3} - x ∣ 2 - 3 x ∣ - 1} :

None

P (x) = x^{4} + 8 x^{2} - 384

f (x) = (2 x^{3} - 3 x^{2} + 5 x)^{3}

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

f (x) = \frac{lo g _{10} ( 2 )}{5 x}

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

What is the f(x)=(4x^2-1)/(6x^3-x|2-3x|-1) ?
The f(x)=(4x^2-1)/(6x^3-x|2-3x|-1) is