What is the f(x)=(ln(x^2))/(sqrt(16-x^4)) ?

The f(x)=(ln(x^2))/(sqrt(16-x^4)) is

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f(x)=(ln(x^2))/(sqrt(16-x^4))

f (x) = \frac{ln ( x ^{2} )}{16 - x ^{4}}

Domain : - 2 < x < 0 or 0 < x < 2

X Intercepts : (1, 0), (- 1, 0)

Asymptotes : Vertical x = 0, x = 2

Extreme Points : None

Interval Notation

Domain : (- 2, 0) \cup (0, 2)

Solution steps

Domain of

\frac{ln ( x ^{2} )}{16 - x ^{4}} : - 2 < x < 0 or 0 < x < 2

Axis interception points of

\frac{ln ( x ^{2} )}{16 - x ^{4}} :

X Intercepts

: (1, 0), (- 1, 0)

Asymptotes of

\frac{ln ( x ^{2} )}{16 - x ^{4}} :

Vertical

: x = 0, x = 2

Extreme Points of

\frac{ln ( x ^{2} )}{16 - x ^{4}} :

None

f (n) = 2^{n + 4} - 2^{n + 3}

f (x) = x^{2} cos (x) d x

P (x) = x^{3} - 39 x - 70

f (x) = - 3 x (x - 5) (x - 3)^{4} (x - 1)

What is the f(x)=(ln(x^2))/(sqrt(16-x^4)) ?
The f(x)=(ln(x^2))/(sqrt(16-x^4)) is