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

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

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

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

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

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

Asymptotes : Horizontal y = 0

Extreme Points : Maximum (\frac{- 3 - 21}{6}, \frac{21 - 3}{6}), Minimum (\frac{- 3 + 21}{6}, - \frac{3 + 21}{6})

Interval Notation

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

Solution steps

Domain of

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

Axis interception points of

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

X Intercepts

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

Y Intercepts

: (0, - 1)

Asymptotes of

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

Horizontal

: y = 0

Extreme Points of

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

Maximum

(\frac{- 3 - 21}{6}, \frac{21 - 3}{6}),

Minimum

(\frac{- 3 + 21}{6}, - \frac{3 + 21}{6})

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

f (x) = - 2 x^{4} + 17 x^{3} - 37 x^{2} + 7 x + 15

f (x) = cos (ln (x)) \cdot tan (ln (x))

f (x) = 5 x^{2} + 10 x + 1

y = - 3 tan (\frac{1}{2} x)

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