An inverse function undoes what the function did. Only a one-one function has one, its domain and range swap places, and its graph is the mirror image of the original in the line \(y = x\).
To find \(f^{-1}\): write \(y = f(x)\), make \(x\) the subject, then swap the letters. The domain of \(f^{-1}\) is the range of \(f\), the range of \(f^{-1}\) is the domain of \(f\), and the graph of \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in the line \(y = x\).
Fig. 1: f(x) = x² - 2 for x ≥ 0 and its inverse f⁻¹(x) = √(x + 2) are mirror images in y = x; they meet on that line at (2, 2)
Section 01
Undo, in the reverse order.
In the lab, pick a function and watch its inverse appear as the reflection in \(y = x\): every point \((a, b)\) on \(y = f(x)\) becomes \((b, a)\). That swap of inputs and outputs is exactly what the algebra does. Remember the inverse only exists when \(f\) is one-one.
\(f^{-1}(x)\) is NOT \(\dfrac{1}{f(x)}\). For \(f(x) = 3x - 5\), \(\dfrac{1}{3x - 5}\) is a reciprocal; the inverse is \(\dfrac{x + 5}{3}\).
Section 02
Inverses of quadratics.
A quadratic is only one-one on one side of its vertex, so the domain is always restricted. Complete the square first, then take a square root, choosing the sign that matches the domain.
Worked example Positive root
\(f(x) = (x - 2)^{2} + 1\) for \(x \ge 2\). Find \(f^{-1}(x)\) and state its domain.
\(y - 1 = (x - 2)^{2}\).
Here \(x \ge 2\), so \(x - 2 \ge 0\) and we take the positive root: \(x - 2 = \sqrt{y - 1}\).
\(x = 2 + \sqrt{y - 1}\). The range of f is \(f(x) \ge 1\).
Writing \(\pm\sqrt{\ }\) in the final answer is wrong: an inverse function gives ONE output. Test your choice: \(g^{-1}\) must output values \(\le \frac{1}{3}\), and \(\frac{1 - \sqrt{x + 5}}{3}\) does.
Stage 1 · Learn
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Find \(f^{-1}(x)\) when \(f(x) = 4x + 1\).
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Which of these functions has an inverse?
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The graph of \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in
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\(f\) has domain \(x \ge 1\) and range \(f(x) \ge 4\). What is the domain of \(f^{-1}\)?
Section 03
Mirror in y = x.
Because the inverse swaps every \((a, b)\) to \((b, a)\), the graphs of \(f\) and \(f^{-1}\) are reflections of each other in \(y = x\). Sketches of both should show the mirror line (Fig. 1). An increasing function meets its inverse only on that line.
Points: if \((0, -2)\) is on \(y = f(x)\), then \((-2, 0)\) is on \(y = f^{-1}(x)\).
Undoing: \(ff^{-1}(x) = x\) and \(f^{-1}f(x) = x\) for every \(x\) in the right domain.
Meeting points of an increasing \(f\) and \(f^{-1}\): solve \(f(x) = x\), which is easier than \(f(x) = f^{-1}(x)\).
Worked example Where a function meets its inverse
\(f(x) = x^{2} - 2\) for \(x \ge 0\). Find the point where \(y = f(x)\) meets \(y = f^{-1}(x)\).
\(f\) is increasing on \(x \ge 0\), so the graphs meet on \(y = x\): solve \(x^{2} - 2 = x\).
\(\sqrt{x^{2}} = x\) because \(x \ge 0\) in the domain of f.
Answer: \(f^{-1}f(x) = x\), so the inverse is correct
Examiner trap
Sketch \(f^{-1}\) only over the RANGE of \(f\). In Fig. 1 the inverse starts at \((-2, 0)\); drawing \(\sqrt{x + 2}\) to the left of \(-2\), or reflecting the whole parabola, earns no marks.
Stage 2 · Exam
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\(f(x) = (x + 2)^{2} - 3\) for \(x \ge -2\). Find \(f^{-1}(x)\) and its domain.