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Popular >

f(x)=sqrt(x)+1/(ln(x))

Solution

f (x) = x + \frac{1}{ln ( x )}

Solution

Domain : 0 < x < 1 or x > 1

Range : f (x) < 0 or f (x) \geq \frac{1}{8 W _{0} ( \frac{1}{2 2} ) ^{4}} + \frac{1}{ln ( \frac{1}{64 W _{0} ( \frac{1}{2 2} ) ^{4}} )}

Intercepts : None

Asymptotes : Vertical x = 1

Extreme Points : Minimum \frac{1}{64 W _{- 1} ( - \frac{1}{2 2} ) ^{4}}, \frac{1}{8 W _{- 1} ( - \frac{1}{2 2} ) ^{4}} + \frac{1}{ln ( \frac{1}{64 W _{- 1} ( - \frac{1}{2 2} ) ^{4}} )}, Maximum \frac{1}{64 W _{0} ( - \frac{1}{2 2} ) ^{4}}, \frac{1}{8 W _{0} ( - \frac{1}{2 2} ) ^{4}} + \frac{1}{ln ( \frac{1}{64 W _{0} ( - \frac{1}{2 2} ) ^{4}} )}, Minimum \frac{1}{64 W _{0} ( \frac{1}{2 2} ) ^{4}}, \frac{1}{8 W _{0} ( \frac{1}{2 2} ) ^{4}} + \frac{1}{ln ( \frac{1}{64 W _{0} ( \frac{1}{2 2} ) ^{4}} )}

+1

Interval Notation

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

Range : (- \infty, 0) \cup \frac{1}{8 W _{0} ( \frac{1}{2 2} ) ^{4}} + \frac{1}{ln ( \frac{1}{64 W _{0} ( \frac{1}{2 2} ) ^{4}} )}, \infty

Solution steps

Domain of

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

Range of

x + \frac{1}{ln ( x )} : f (x) < 0 or f (x) \geq \frac{1}{8 W _{0} ( \frac{1}{2 2} ) ^{4}} + \frac{1}{ln ( \frac{1}{64 W _{0} ( \frac{1}{2 2} ) ^{4}} )}

Axis interception points of

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

None

Asymptotes of

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

Vertical

: x = 1

Extreme Points of

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

Minimum

\frac{1}{64 W _{- 1} ( - \frac{1}{2 2} ) ^{4}}, \frac{1}{8 W _{- 1} ( - \frac{1}{2 2} ) ^{4}} + \frac{1}{ln ( \frac{1}{64 W _{- 1} ( - \frac{1}{2 2} ) ^{4}} )},

Maximum

\frac{1}{64 W _{0} ( - \frac{1}{2 2} ) ^{4}}, \frac{1}{8 W _{0} ( - \frac{1}{2 2} ) ^{4}} + \frac{1}{ln ( \frac{1}{64 W _{0} ( - \frac{1}{2 2} ) ^{4}} )},

Minimum

\frac{1}{64 W _{0} ( \frac{1}{2 2} ) ^{4}}, \frac{1}{8 W _{0} ( \frac{1}{2 2} ) ^{4}} + \frac{1}{ln ( \frac{1}{64 W _{0} ( \frac{1}{2 2} ) ^{4}} )}

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Frequently Asked Questions (FAQ)

What is the f(x)=sqrt(x)+1/(ln(x)) ?
The f(x)=sqrt(x)+1/(ln(x)) is