What is the f(x)=ln(x^2-1) ?

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f(x)=ln(x^2-1)

f (x) = ln (x^{2} - 1)

Domain : x < - 1 or x > 1

Range : - \infty < f (x) < \infty

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

Asymptotes : Vertical x = - 1, x = 1

Extreme Points : None

Interval Notation

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

Range : (- \infty, \infty)

Solution steps

Domain of

ln (x^{2} - 1) : x < - 1 or x > 1

Range of

ln (x^{2} - 1) : - \infty < f (x) < \infty

Axis interception points of

ln (x^{2} - 1) :

X Intercepts

: (2, 0), (- 2, 0)

Asymptotes of

ln (x^{2} - 1) :

Vertical

: x = - 1, x = 1

Extreme Points of

ln (x^{2} - 1) :

None

\frac{d}{d x} (\frac{x ^{2} - 6 x}{( x - 3 ) ^{2}})

\int \frac{3}{( x - 2 ) ^{2}} d x

(12, - 16), (- 9, 8)

\int (tan (x) (6 cos (x) - 6 sec (x))) d x

y^{^{''}} - y^{^{'}} - 6 y = - 4 e^{2} t, y (0) = 0, y^{^{'}} (0) = 0