domain of f(x)=ln(1/x-1/(1-x)) — step-by-step solution

We save the definition:

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

1) Let's evaluate the domain of

\ln (\frac{1}{x} - \frac{1}{1 - x})

For the expression

\ln (\frac{1}{x} - \frac{1}{1 - x})

to exist, you must check:

\frac{1}{x} - \frac{1}{1 - x} > 0

For the expression

\frac{1}{x}

to be defined, we must have:

x \neq 0

For the expression

- \frac{1}{1 - x}

to exist, you must check:

1 - x \neq 0

2) Let's solve in

ℝ

\frac{1}{x} - \frac{1}{1 - x} > 0

3) Let's add fractions:

\frac{1}{x} + \frac{1}{- 1 + x} > 0

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{1}{x} + \frac{1}{- 1 + x}

\frac{(- 1 + x) + x}{x \cdot (- 1 + x)}

\frac{(- 1 + x) + x}{x \cdot (- 1 + x)} > 0

ⓘ Like terms are grouped by adding their coefficients:

x + x

2 x

\frac{- 1 + 2 x}{x \cdot (- 1 + x)} > 0

We study the sign of the numerator

⇥The expression

- 1 + 2 x

is zero at:

⇥

x = \frac{1}{2}

⇥Its coefficient of

x

2

, being positive, it is negative before and positive after

\frac{1}{2}

We study the sign of the denominator, which must not be zero

Let's solve in

ℝ

x \cdot (- 1 + x) > 0

We study the sign of each factor of the product

⇥The expression

x

is zero at

x = 0

(forbidden value), it is negative before and positive after

⇥The expression

- 1 + x

is zero at

x = 1

(forbidden value), it is negative before and positive after

4) Let's draw up the sign table:

Solution found:

x \in (0, \frac{1}{2}) \cup (1, +∞)

5) Let's solve in

ℝ

1 - x \neq 0

ⓘ We add

x

to both sides

1 \neq x

Solution obtained:

x \neq 1

Let's build the table of the different terms to determine the global solution:

▶Domain of definition:

x \in (0, \frac{1}{2}) \cup (1, +∞)