What is the f(x)=([(ln(x))^2+1])/(ln(x)) ?

The f(x)=([(ln(x))^2+1])/(ln(x)) is

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f(x)=([(ln(x))^2+1])/(ln(x))

f (x) = \frac{[ ( ln ( x ) ) ^{2} + 1 ]}{ln ( x )}

Domain : 0 < x < 1 or x > 1

Range : f (x) \leq - 2 or f (x) \geq 2

Intercepts : None

Asymptotes : Vertical x = 0, x = 1

Extreme Points : Maximum (\frac{1}{e}, - 2), Minimum (e, 2)

Interval Notation

Domain : (0, 1) \cup (1, \infty)

Range : (- \infty, - 2] \cup [2, \infty)

Solution steps

Domain of

\frac{ln ^{2} ( x ) + 1}{ln ( x )} : 0 < x < 1 or x > 1

Range of

\frac{ln ^{2} ( x ) + 1}{ln ( x )} : f (x) \leq - 2 or f (x) \geq 2

Axis interception points of

\frac{ln ^{2} ( x ) + 1}{ln ( x )} :

None

Asymptotes of

\frac{ln ^{2} ( x ) + 1}{ln ( x )} :

Vertical

: x = 0, x = 1

Extreme Points of

\frac{ln ^{2} ( x ) + 1}{ln ( x )} :

Maximum

(\frac{1}{e}, - 2),

Minimum

(e, 2)

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