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

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

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

domain

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

0 < x < 1 or x > 1

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \leq - 2 or x \geq 0

Find positive values for logs:

x > 0

Find undefined (singularity) points:

x = 1

Combine real regions and undefined points for final function domain

0 < x < 1 or x > 1

f (x) = \frac{30000}{100 - x}

f (x) = ∣ x - 1 ∣ + ∣ x + 1 ∣

- 4 - 2 x + 1 - 2

f (x) = \frac{3 x}{4 x - 7}

f (x) = x - 3 + 2 x

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