Functions, graphs and inverses, question 16

Topic: Functions, graphs and inverses
Difficulty: Challenging
Suggested time: 7 min
Marks: 4
Theory: UBC Math 100 §4.6 CLP-1 §0.6Key results

Let $f(x) = \log\!\left(\dfrac{1 + x}{1 - x}\right)$ . Find the domain of $f$ , show that $f$ is odd, and find a formula for $f^{-1}(y)$ .

Your answer

Theory UBC Math 100 §4.6 CLP-1 §0.6Key results

Worked solution

Strategy

The three parts use three different tools: an inequality for the domain, a log identity for oddness, and exponentiation for the inverse. Do them in that order; the domain tells you what the inverse’s range must be.

Working

$\frac{1 + x}{1 - x} > 0 \iff -1 < x < 1$
Numerator and denominator must have the same sign. Both are positive exactly on (−1, 1); both negative would need x < −1 and x > 1 at once, which never happens.
$f(-x) = \log\!\left(\frac{1 - x}{1 + x}\right) = -\log\!\left(\frac{1 + x}{1 - x}\right) = -f(x)$
Swapping x for −x flips the fraction, and log(1/u) = −log u.
$e^{y} = \frac{1 + x}{1 - x} \quad\Longrightarrow\quad e^{y} - x e^{y} = 1 + x$
Exponentiate, then clear the denominator.
$x\bigl(e^{y} + 1\bigr) = e^{y} - 1 \quad\Longrightarrow\quad f^{-1}(y) = \frac{e^{y} - 1}{e^{y} + 1}$
Collect the x terms. The result lies in (−1, 1) for every y, matching the domain of f.

Answer

\text{Domain } (-1, 1);\quad f \text{ odd};\quad f^{-1}(y) = \frac{e^{y} - 1}{e^{y} + 1}

What earns the marks

Domain: (1 + x)/(1 − x) > 0, giving −1 < x < 1, with both endpoints excluded.
Oddness: f(−x) = log((1 − x)/(1 + x)) = −log((1 + x)/(1 − x)) = −f(x), using log(1/u) = −log u.
Inverse: solve e^y = (1 + x)/(1 − x) for x to get f⁻¹(y) = (e^y − 1)/(e^y + 1), defined for every real y.

Check

At y = log 3 the inverse gives (3 − 1)/(3 + 1) = 1/2, and f(1/2) = log(1.5/0.5) = log 3, so the two undo each other.

Where this goes wrong

Giving the domain as x ≠ 1, forgetting that the argument of a logarithm must be positive, not merely defined. For x < −1 the fraction is negative and f does not exist.

Try a similar question

Let

f(x) = \dfrac{3x - 1}{x + 2}

. Find

f^{-1}(1)

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Functions, graphs and inverses practice question 16