What is the domain of f(x)=-1/(x(4+ln(x))^2) ?

The domain of f(x)=-1/(x(4+ln(x))^2) is 0 1/(e^4)

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domain of f(x)=-1/(x(4+ln(x))^2)

domain

f (x) = - \frac{1}{x ( 4 + ln ( x ) ) ^{2}}

0 < x < \frac{1}{e ^{4}} or x > \frac{1}{e ^{4}}

Interval Notation

(0, \frac{1}{e ^{4}}) \cup (\frac{1}{e ^{4}}, \infty)

Solution steps

Find positive values for logs:

x > 0

Find undefined (singularity) points:

x = \frac{1}{e ^{4}}

Combine real regions and undefined points for final function domain

0 < x < \frac{1}{e ^{4}} or x > \frac{1}{e ^{4}}

f (x) = \frac{2 x + 1}{x - 5}

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

e^{4 x + 3} + 2

f (x) = - x - 1, x < 0

x^{2} + 2 x + 1

What is the domain of f(x)=-1/(x(4+ln(x))^2) ?
The domain of f(x)=-1/(x(4+ln(x))^2) is 0<x< 1/(e^4)\lor x> 1/(e^4)