An inverse function takes every output of \(f\) back to the input it came from. It only exists when \(f\) is one-one, its domain and range are those of \(f\) swapped, and its graph is the reflection of \(y = f(x)\) in the line \(y = x\).
Only a one-one function has an inverse. 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) = eˣ + 1 and its inverse f⁻¹(x) = ln(x - 1) are mirror images in y = x. The intercept (0, 2) becomes (2, 0), and the asymptote y = 1 becomes x = 1
Section 01
One-one or many-one?
An inverse must send each output back to ONE input. If two different inputs give the same output, \(f\) is many-one and the inverse would not know which input to return, so \(f\) has no inverse. On a graph, a function is one-one when every horizontal line meets it at most once. A quadratic on all real numbers always fails, but restricting its domain to one side of the vertex makes it one-one.
Explain in words: give two inputs with the same output, for example \(f(0) = f(4)\), and say the function is many-one, so it has no inverse.
Restrict a quadratic: complete the square to find the vertex \((p, q)\), then use \(x \ge p\) or \(x \le p\).
Exponential and log functions such as \(e^{2x}\) and \(\ln x\) are already one-one on their natural domains.
Worked example Why there is no inverse
The function f is defined by \(f(x) = x^{2} - 4x\) for \(x \in \mathbb{R}\). Explain why f does not have an inverse.
\(f(0) = 0\) and \(f(4) = 16 - 16 = 0\).
Two different inputs, 0 and 4, give the same output, 0, so f is many-one.
An inverse would have to send 0 back to both 0 and 4, which a function cannot do.
Answer: f is many-one (for example \(f(0) = f(4) = 0\)), so it has no inverse
Worked example Restrict, then invert
The function g is defined by \(g(x) = x^{2} - 4x\) for \(x \ge 2\). Find \(g^{-1}(x)\) and state its domain.
Complete the square: \(g(x) = (x - 2)^{2} - 4\). The vertex is \((2, -4)\), so on \(x \ge 2\) the graph only rises and g is one-one.
Write \(y = (x - 2)^{2} - 4\), so \((x - 2)^{2} = y + 4\).
Here \(x \ge 2\), so \(x - 2 \ge 0\) and we take the positive root: \(x = 2 + \sqrt{y + 4}\).
The range of g is \(g(x) \ge -4\), which becomes the domain of \(g^{-1}\).
\(h(x) = (x + 3)^{2} + 1\) for \(x \ge k\). Find the least value of \(k\) for which \(h^{-1}\) exists.
The vertex is at \(x = -3\). For \(x \ge k\) to stay on one side of it, \(k \ge -3\), so the least value is \(k = -3\).
Examiner trap
'It is a quadratic' is not an explanation. The mark is for showing that two inputs share an output (or saying a horizontal line meets the graph twice). And in the final answer write only one sign of the root: \(\pm\sqrt{\ }\) would give two outputs, so it is not a function.
Section 02
Find the inverse, with the right notation.
The method is always the same: put \(y = f(x)\), make \(x\) the subject, then replace \(y\) by \(x\) and write the answer as \(f^{-1}(x) = \ldots\). For exponentials, take logs; for logarithms, raise \(e\) to both sides. Finish by stating the domain of \(f^{-1}\), which is the range of \(f\).
Worked example Exponential
The function f is defined by \(f(x) = e^{2x}\) for \(x \in \mathbb{R}\). Find \(f^{-1}(x)\) and state its domain.
Write \(y = e^{2x}\).
Take natural logs: \(\ln y = 2x\), so \(x = \dfrac{1}{2}\ln y\).
Swap the letters: \(f^{-1}(x) = \dfrac{1}{2}\ln x\).
\(e^{2x}\) is always positive and takes every positive value, so the range of f is \(f(x) > 0\).
Answer: \(f^{-1}(x) = \dfrac{1}{2}\ln x\) for \(x > 0\)
Worked example Logarithmic
The function g is defined by \(g(x) = 3 + \ln(2x - 1)\) for \(x > \dfrac{1}{2}\). Find \(g^{-1}(x)\) and state its domain and range.
Write \(y = 3 + \ln(2x - 1)\), so \(\ln(2x - 1) = y - 3\).
Raise \(e\) to both sides: \(2x - 1 = e^{y - 3}\), so \(x = \dfrac{1 + e^{y - 3}}{2}\).
\(\ln(2x - 1)\) takes every real value for \(x > \frac{1}{2}\), so the range of g is all real numbers: the domain of \(g^{-1}\).
The range of \(g^{-1}\) is the domain of g, \(g^{-1}(x) > \frac{1}{2}\).
Answer: \(g^{-1}(x) = \dfrac{1 + e^{x - 3}}{2}\) for \(x \in \mathbb{R}\), with range \(g^{-1}(x) > \dfrac{1}{2}\)
Your turn
\(f(x) = 2e^{x} - 3\) for \(x \in \mathbb{R}\). Find \(f^{-1}(x)\) and state its domain.
\(y + 3 = 2e^{x}\), so \(x = \ln\left(\dfrac{y + 3}{2}\right)\). \(f^{-1}(x) = \ln\left(\dfrac{x + 3}{2}\right)\) for \(x > -3\), because \(2e^{x} > 0\) gives \(f(x) > -3\).
Examiner trap
\(f^{-1}(x)\) is NOT \(\dfrac{1}{f(x)}\). For \(f(x) = e^{2x}\), \(\dfrac{1}{e^{2x}} = e^{-2x}\) is the reciprocal; the inverse is \(\dfrac{1}{2}\ln x\). Also, \(\ln y = 2x\) gives \(x = \frac{1}{2}\ln y\), not \(\ln\frac{y}{2}\).
Stage 1 · Learn
Check your understanding
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Why does \(f(x) = x^{2}\) for \(x \in \mathbb{R}\) not have an inverse?
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\(f(x) = e^{2x}\) for \(x \in \mathbb{R}\). Find \(f^{-1}(x)\).
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\(f(x) = e^{x} + 1\) for \(x \in \mathbb{R}\). What is the domain of \(f^{-1}\)?
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The point \((2, 7)\) lies on \(y = f(x)\). Which point must lie on \(y = f^{-1}(x)\)?
Section 03
Reflect in y = x.
In the lab, choose a function and watch its inverse appear as its mirror image. Because the inverse swaps every point \((a, b)\) on \(y = f(x)\) to \((b, a)\), the two graphs are reflections of each other in the line \(y = x\). Intercepts swap axes and asymptotes swap too: a horizontal asymptote \(y = c\) of \(f\) becomes a vertical asymptote \(x = c\) of \(f^{-1}\) (Fig. 1).
Worked example Sketch a function and its inverse
\(f(x) = e^{x} + 1\) for \(x \in \mathbb{R}\). On one diagram sketch \(y = f(x)\) and \(y = f^{-1}(x)\), showing the line \(y = x\), the intercepts and the asymptotes.
\(y = f(x)\): \(f(0) = 2\), so the \(y\)-intercept is \((0, 2)\). As \(x \to -\infty\), \(e^{x} \to 0\), so \(y = 1\) is an asymptote. There is no \(x\)-intercept.
Reflect in \(y = x\): \((0, 2)\) becomes \((2, 0)\) and \(y = 1\) becomes \(x = 1\).
Check with algebra: \(y - 1 = e^{x}\), so \(f^{-1}(x) = \ln(x - 1)\) for \(x > 1\), and \(\ln(2 - 1) = 0\).
Draw \(f^{-1}\) only for \(x > 1\), rising slowly and dropping steeply towards \(x = 1\).
Answer: As in Fig. 1: \(f\) through \((0, 2)\) above \(y = 1\); \(f^{-1}(x) = \ln(x - 1)\) through \((2, 0)\) to the right of \(x = 1\)
Worked example Where a function meets its inverse
\(f(x) = x^{2} - 6\) for \(x \ge 0\). Find the point where the graphs of \(y = f(x)\) and \(y = f^{-1}(x)\) meet.
\(f\) is increasing on \(x \ge 0\), so the two graphs meet on their mirror line \(y = x\): solve \(f(x) = x\).
\(x = -2\) is outside the domain \(x \ge 0\), so reject it.
Answer: \((3,\ 3)\)
Examiner trap
Sketch \(f^{-1}\) only over the RANGE of \(f\), and keep both curves on the correct side of their asymptotes. Reflecting in the \(x\)-axis or \(y\)-axis instead of \(y = x\) is the most common lost mark: check that \((0, 2)\) has become \((2, 0)\), not \((0, -2)\).
Stage 2 · Exam
Exam-style questions
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The function f is defined by \(f(x) = 2x^{2} - 8x + 5\) for \(x \ge k\). Find the least value of \(k\) for which f has an inverse.
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The function g is defined by \(g(x) = \ln(3x + 2)\) for \(x > -\dfrac{2}{3}\). Find \(g^{-1}(x)\).
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The function h is defined by \(h(x) = 4 + 3e^{2x}\) for \(x \in \mathbb{R}\). State the domain of \(h^{-1}\).
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The function f is defined by \(f(x) = (x + 1)^{2} + 3\) for \(x \le -1\). Find \(f^{-1}(x)\).
Skill unlocked
Inverses found, explained and reflected.
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