What is the y=cos(2x)+ln(x^2) ?

The y=cos(2x)+ln(x^2) is Domain: x 0 Intercepts: None Asymptotes: Vertical x=0

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y=cos(2x)+ln(x^2)

y = cos (2 x) + ln (x^{2})

Domain : x < 0 or x > 0

Intercepts : None

Asymptotes : Vertical x = 0

Interval Notation

Domain : (- \infty, 0) \cup (0, \infty)

Solution steps

Domain of

cos (2 x) + ln (x^{2}) : x < 0 or x > 0

Axis interception points of

cos (2 x) + ln (x^{2}) :

None

Asymptotes of

cos (2 x) + ln (x^{2}) :

Vertical

: x = 0

p (y) = 3 x^{3} - 6 x^{2} + x - 3

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

f (x) = \frac{1}{( e ^{x} + x )}

f (x) = - 3 x^{2} - 3 x + 6

g (x) = \frac{- x ^{2}}{4 + 7 ( - 2.4 )}

What is the y=cos(2x)+ln(x^2) ?
The y=cos(2x)+ln(x^2) is Domain: x<0\lor x>0<br/>Intercepts: None<br/>Asymptotes: Vertical x=0