P3.1 Algebraic fractions and functions
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P3 (Pure Mathematics 3), specification sections 1.1 and 1.2. The modulus function and combined graph transformations are the next topics in P3; partial fractions belong to P4.
What this page covers
7 points
- Simplifying algebraic fractions by factorising and cancelling common factors
- Multiplying, dividing, adding and subtracting algebraic fractions with linear or quadratic denominators
- Algebraic division of an improper fraction by a linear or quadratic expression
- Functions: domain, range, one-one and many-one mappings, and why a one-many mapping is not a function
- Composite functions: \(fg(x)\) means apply \(g\) first, then \(f\)
- Inverse functions, their domain and range, and the reflection of the graph in \(y = x\)
- Restricting a domain so that an inverse exists
In one minute
6 points
- Algebraic fractions follow the same rules as number fractions provided you factorise first and cancel only factors, and a function is a rule together with its domain, so every range, composite and inverse depends on which inputs are allowed.
- Combine: for \(\times\) and \(\div\), factorise and cancel (\(\div\) means \(\times\) by the reciprocal); for \(+\) and \(-\), use the lowest common denominator, then simplify the numerator.
- Divide: if the numerator's degree is at least the denominator's, write \(\frac{N(x)}{D(x)} \equiv Q(x) + \frac{R(x)}{D(x)}\) with the degree of \(R\) less than the degree of \(D\).
- Function: every input in the domain has exactly one output. The range is the set of outputs, written in terms of \(f(x)\).
- Composite: \(fg(x) = f(g(x))\), so \(g\) acts first.
- Inverse: exists only for a one-one function. Domain of \(f^{-1}\) = range of \(f\); its graph is the reflection of \(y = f(x)\) in \(y = x\).
Contents
13 sections
- How do I simplify and combine algebraic fractions?
- What do I do when the numerator's degree is too high?
- What are the domain and range of a function?
- How do I form a composite function?
- How do I find an inverse function?
- How do I restrict a domain so that an inverse exists?
- How do these ideas combine in a full question?
- Exam tips for P3.1
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
How do I simplify and combine algebraic fractions?
Nearly every later topic in P3 and P4 produces fractions with \(x\) in them: the quotient rule, partial fractions, parametric gradients. Handling them confidently rests on one idea you already use with numbers: \(\frac{6}{8} = \frac{2 \times 3}{2 \times 4} = \frac34\) works because 2 is a factor of both top and bottom.
| Operation | Method | Example |
|---|---|---|
| Simplify | factorise top and bottom fully; cancel common factors | \(\frac{x^2 - 4}{x^2 + 2x} = \frac{(x-2)(x+2)}{x(x+2)} = \frac{x-2}{x}\) |
| Multiply | factorise everything, cancel across the fractions, then multiply | \(\frac{3}{x+1} \times \frac{x^2-1}{6} = \frac{x-1}{2}\) |
| Divide | multiply by the reciprocal of the second fraction | \(\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}\) |
| Add or subtract | use the lowest common denominator; expand and collect the numerator; check whether it now factorises and cancels | \(\frac1x + \frac{2}{x+1} = \frac{3x+1}{x(x+1)}\) |
(a) Simplify fully \(\dfrac{2x^2 + 5x - 3}{4x^2 - 1}\). (b) Simplify fully \(\dfrac{x^2 - 9}{x^2 + 2x - 15} \div \dfrac{2x + 6}{x + 5}\).
(a) Factorise top and bottom (the bottom is a difference of two squares): \(\frac{(x+3)(2x-1)}{(2x-1)(2x+1)}\). Cancel the common factor \((2x - 1)\). \(\frac{x+3}{2x+1}\)
(b) Turn the division into multiplication by the reciprocal and factorise everything: \(\frac{(x-3)(x+3)}{(x-3)(x+5)} \times \frac{x+5}{2(x+3)}\). Cancel \((x - 3)\), \((x + 3)\) and \((x + 5)\). \(\frac12\)
The answer to (b) is a constant: every factor cancels. It holds for every \(x\) for which the original expression is defined (\(x \ne 3, -3, -5\)).
Cancelling terms instead of factors: \(\frac{x+3}{x+5}\) is not \(\frac35\), and \(\frac{x^2+4}{x^2}\) is not 4. You can only cancel something that multiplies the whole numerator and the whole denominator.
Show that \(\dfrac{2}{x - 3} - \dfrac{x + 7}{x^2 - x - 6}\) can be written as \(\dfrac{1}{x + 2}\).
Factorise the quadratic denominator to find the lowest common denominator: \(x^2 - x - 6 = (x - 3)(x + 2)\), so the common denominator is \((x - 3)(x + 2)\).
Rewrite the first fraction over it and subtract: \[\frac{2(x+2) - (x+7)}{(x-3)(x+2)} = \frac{2x + 4 - x - 7}{(x-3)(x+2)} = \frac{x - 3}{(x-3)(x+2)}\]
Cancel the common factor. \(= \frac{1}{x+2}\) as required
Put the subtracted numerator in brackets: \(-(x + 7) = -x - 7\). Losing that bracket is the classic slip.
What do I do when the numerator's degree is too high?
Find \(Q\) and \(R\) by long division or by comparing coefficients in \(N(x) \equiv Q(x)D(x) + R(x)\).
A linear denominator
For \(\frac{3x + 5}{x + 1}\), write the top in terms of the bottom: \(3x + 5 = 3(x + 1) + 2\), so \(\frac{3x+5}{x+1} = 3 + \frac{2}{x+1}\). The divided form shows the behaviour at once. As \(x \to \pm\infty\) the fraction part tends to 0, so \(y \to 3\), a horizontal asymptote; and the fraction part is never 0, so \(y = 3\) is never reached. That will be useful when finding a range.
A quadratic denominator
Long division works exactly as for a linear divisor: divide the leading terms, multiply back, subtract, repeat. Keep a 0 in place of any missing power in the divisor or the dividend (for example \(x^2 + 0x + 2\)), and stop as soon as what is left has degree less than 2: that linear (or constant) expression is the remainder.
Express \(\dfrac{2x^4 - x^3 + 7x^2 - x + 2}{x^2 + 2}\) in the form \(ax^2 + bx + c + \dfrac{dx + e}{x^2 + 2}\), where \(a, b, c, d\) and \(e\) are constants.
\(2x^4 \div x^2 = 2x^2\); subtract \(2x^2(x^2 + 2) = 2x^4 + 4x^2\), leaving \(-x^3 + 3x^2 - x + 2\).
\(-x^3 \div x^2 = -x\); subtract \(-x(x^2 + 2) = -x^3 - 2x\), leaving \(3x^2 + x + 2\).
\(3x^2 \div x^2 = 3\); subtract \(3(x^2 + 2) = 3x^2 + 6\), leaving \(x - 4\), which has degree \(1 < 2\), so it is the remainder.
\(2x^2 - x + 3 + \frac{x - 4}{x^2 + 2}\) (\(a = 2,\ b = -1,\ c = 3,\ d = 1,\ e = -4\))
Quick check at \(x = 1\): left \(= \frac{2 - 1 + 7 - 1 + 2}{3} = 3\); right \(= 2 - 1 + 3 + \frac{-3}{3} = 3\) ✓
Comparing coefficients is a reliable alternative: expand \((ax^2 + bx + c)(x^2 + 2) + dx + e\) and match \(x^4\), \(x^3\), \(x^2\), \(x\) and the constant one at a time, starting from the highest power.
What are the domain and range of a function?
- Notation: \(f(x) = 3x - 1\), or \(f : x \mapsto 3x - 1\), together with a domain such as \(x \in \mathbb{R}\) or \(x > 2\).
- The domain is part of the function: the same rule with a different domain is a different function.
- The range is a set of outputs, so write it with \(f(x)\) (or \(y\)): "\(f(x) \ge 3\)", not "\(x \ge 3\)".
- One-one: different inputs always give different outputs. Many-one: at least two inputs share an output. Both are functions.
- One-many (one input, several outputs) is not a function.
- Graph test: a vertical line meets the graph of a function at most once; a horizontal line meets the graph of a one-one function at most once.
Finding a range
- Sketch the graph over the domain only, then read off the \(y\)-values it covers.
- For a quadratic, complete the square to find the vertex.
- For a fraction, divide first: asymptotes show values that are approached but never reached.
- Check the end points of the domain: is the end value included (\(\le, \ge\)) or not (\(<, >\))?
(a) \(f(x) = x^2 - 4x + 7\), \(x \in \mathbb{R}\). Find the range of \(f\) and explain whether \(f\) is one-one or many-one. (b) \(g(x) = 3 + \dfrac{2}{x + 1}\), \(x \ge 0\). Find the range of \(g\).
(a) Complete the square: \(f(x) = (x - 2)^2 + 3\), and \((x - 2)^2 \ge 0\) with equality at \(x = 2\). \(f(x) \ge 3\) Two inputs with the same output: \(f(1) = f(3) = 4\), so \(f\) is many-one.
(b) Use the domain. For \(x \ge 0\), \(x + 1 \ge 1\), so \(0 < \frac{2}{x+1} \le 2\). The largest value is \(g(0) = 5\); as \(x \to \infty\), \(g(x) \to 3\) but never equals 3. \(3 < g(x) \le 5\)
How do I form a composite function?
\(fg\) can only be formed if every output of \(g\) is an allowed input of \(f\): the range of \(g\) must lie inside the domain of \(f\). The domain of \(fg\) is the domain of \(g\).
The functions \(f\) and \(g\) are defined by \(f(x) = 3x - 1\), \(x \in \mathbb{R}\), and \(g(x) = x^2 + 2\), \(x \in \mathbb{R}\). (a) Find \(fg(2)\) and \(gf(2)\). (b) Find \(fg(x)\) and \(gf(x)\). (c) Solve \(fg(x) = gf(x)\), giving exact answers.
(a) Innermost function first: \(fg(2) = f(6) = 17\); \(gf(2) = g(5) = 27\). 17 and 27
(b) Substitute the whole of the inner function: \(fg(x) = 3(x^2 + 2) - 1 = 3x^2 + 5\); \(gf(x) = (3x - 1)^2 + 2 = 9x^2 - 6x + 3\). \(fg(x) = 3x^2 + 5,\ gf(x) = 9x^2 - 6x + 3\)
(c) Equate and rearrange: \(9x^2 - 6x + 3 = 3x^2 + 5 \Rightarrow 6x^2 - 6x - 2 = 0 \Rightarrow 3x^2 - 3x - 1 = 0\). Quadratic formula: \(x = \frac{3 \pm \sqrt{9 + 12}}{6}\). \(x = \frac{3 \pm \sqrt{21}}{6}\) (\(\approx 1.26\) or \(-0.264\))
Reading \(fg(x)\) from left to right and applying \(f\) first. The function written next to \(x\) acts first, exactly as in \(f(g(x))\). Another slip: multiplying the two rules, \(f(x) \times g(x)\), which is a product, not a composite.
How do I find an inverse function?
- Only a one-one function has an inverse: if two inputs shared an output, the reverse mapping would be one-many.
- Method: write \(y = f(x)\); rearrange to make \(x\) the subject; then write \(f^{-1}(x)\) by replacing \(y\) with \(x\).
- Domain of \(f^{-1}\) = range of \(f\), and range of \(f^{-1}\) = domain of \(f\). Always state the domain of the inverse.
- Graph: \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in the line \(y = x\): the point \((a, b)\) becomes \((b, a)\), and a vertical asymptote \(x = k\) becomes a horizontal asymptote \(y = k\).
\(f(x) = \dfrac{2x + 3}{x - 1}\), \(x > 1\). (a) Show that \(f(x) = 2 + \dfrac{5}{x - 1}\) and hence find the range of \(f\). (b) Find \(f^{-1}(x)\), stating its domain. (c) Solve \(f(x) = f^{-1}(x)\).
(a) Write the numerator in terms of the denominator: \(2x + 3 = 2(x - 1) + 5\), so \(f(x) = 2 + \frac{5}{x-1}\). For \(x > 1\), \(x - 1\) takes every positive value, so \(\frac{5}{x-1}\) takes every positive value. \(f(x) > 2\)
(b) Let \(y = \frac{2x+3}{x-1}\) and make \(x\) the subject: \(y(x - 1) = 2x + 3 \Rightarrow xy - 2x = y + 3 \Rightarrow x = \frac{y+3}{y-2}\). \(f^{-1}(x) = \frac{x+3}{x-2},\ x > 2\) (domain = range of \(f\))
(c) Equate (both sides are defined for \(x > 2\)) and cross-multiply: \((2x + 3)(x - 2) = (x + 3)(x - 1) \Rightarrow 2x^2 - x - 6 = x^2 + 2x - 3 \Rightarrow x^2 - 3x - 3 = 0\). So \(x = \frac{3 \pm \sqrt{21}}{2}\); only \(\frac{3 + \sqrt{21}}{2} \approx 3.79\) satisfies \(x > 2\). \(x = \frac{3 + \sqrt{21}}{2}\)
The solution lies on \(y = x\): here \(f(x) = x\) gives the same equation \(x^2 - 3x - 3 = 0\).
How do I restrict a domain so that an inverse exists?
A many-one function has no inverse, but a restricted version of it can. Choose a domain on which the graph is always increasing or always decreasing; then every horizontal line meets it at most once.
- For a quadratic, split at the vertex: the domain \(x \ge\) (vertex \(x\)-value) or \(x \le\) (vertex \(x\)-value).
- The restricted domain decides the sign when you take a square root in the inverse: the outputs of \(f^{-1}\) must lie in the restricted domain of \(f\).
\(f(x) = x^2 - 6x + 5\). (a) Write \(f(x)\) in the form \((x - p)^2 + q\). (b) Explain why \(f\), with domain \(x \in \mathbb{R}\), has no inverse. (c) With the domain \(x \ge k\), \(f\) is one-one. State the least value of \(k\). (d) For that domain, find \(f^{-1}(x)\) and state its domain and range.
(a) Complete the square. \(f(x) = (x - 3)^2 - 4\)
(b) \(f(2) = f(4) = -3\), so \(f\) is many-one on \(\mathbb{R}\) and has no inverse.
(c) The vertex is at \(x = 3\); to the right of it the graph only increases. \(k = 3\)
(d) Let \(y = (x - 3)^2 - 4\) and rearrange: \((x - 3)^2 = y + 4 \Rightarrow x - 3 = +\sqrt{y + 4}\) (positive root, since \(x \ge 3\)). \(f^{-1}(x) = 3 + \sqrt{x + 4},\ x \ge -4\); range \(f^{-1}(x) \ge 3\)
The domain of \(f^{-1}\) is the range of the restricted \(f\), which is \(f(x) \ge -4\).
Writing \(f^{-1}(x) = 3 \pm \sqrt{x + 4}\). A function gives one output, and the restricted domain \(x \ge 3\) tells you which sign: the outputs of \(f^{-1}\) must be at least 3, so take \(+\). With the domain \(x \le 3\) instead you would take \(-\).
How do these ideas combine in a full question?
Longer questions chain the skills on this page. Simplify the fraction first; it makes the range, the inverse and any composite much easier. Use the divided form (constant + proper fraction) to find the range from the domain. Each part usually feeds the next: "hence" means use the result you have just found.
The function \(f\) is defined by \(f(x) = \dfrac{x^2 - x - 2}{x^2 - 4}\), \(x > 2\). (a) Show that \(f(x) = \dfrac{x+1}{x+2}\) and hence that \(f(x) = 1 - \dfrac{1}{x+2}\). (b) Find the range of \(f\). (c) Find \(f^{-1}(x)\), stating its domain. (d) The function \(g\) is defined by \(g(x) = 4x - 3\), \(x \in \mathbb{R}\). Show that \(gf(x) = \dfrac{x - 2}{x + 2}\) and solve \(gf(x) = \frac13\).
(a) Factorise and cancel: \(\frac{(x-2)(x+1)}{(x-2)(x+2)} = \frac{x+1}{x+2}\) (allowed, as \(x - 2 \ne 0\) for \(x > 2\)). Then \(x + 1 = (x + 2) - 1\), so \(f(x) = 1 - \frac{1}{x+2}\).
(b) \(x > 2 \Rightarrow x + 2 > 4 \Rightarrow 0 < \frac{1}{x+2} < \frac14\). \(\frac34 < f(x) < 1\)
(c) \(y = \frac{x+1}{x+2} \Rightarrow xy + 2y = x + 1 \Rightarrow x(y - 1) = 1 - 2y\). \(f^{-1}(x) = \frac{2x - 1}{1 - x},\ \frac34 < x < 1\)
(d) \(f\) acts first, then \(g\): \(gf(x) = 4 \cdot \frac{x+1}{x+2} - 3 = \frac{4x + 4 - 3x - 6}{x+2} = \frac{x-2}{x+2}\) ✓. Solve \(\frac{x-2}{x+2} = \frac13\): \(3(x - 2) = x + 2 \Rightarrow 2x = 8\). \(x = 4\) (which satisfies \(x > 2\))
Check (c): \(f(4) = \frac56\) and \(f^{-1}\left(\frac56\right) = \frac{2/3}{1/6} = 4\) ✓
After solving an equation involving a function, check each answer against the domain. A root that falls outside the domain must be rejected, with a reason: "\(x = -0.79\) is rejected because \(x > 1\)".
Exam tips for P3.1
Whether you are simplifying, multiplying or adding, factorise every numerator and denominator first. The common factors and the lowest common denominator are then visible, and the final answer is already in its simplest form.
For "express in the form …" questions, show each subtraction line, or set out the comparison of coefficients clearly. Then state every constant asked for, and check by substituting a simple value such as \(x = 0\) or 1 into both sides.
Write a range as an inequality in \(f(x)\), and decide carefully between \(<\) and \(\le\) at each end: a value reached at an included domain end point is included; a value approached along an asymptote is not.
Finding the rule for \(f^{-1}\) is only part of the answer. Its domain is the range of \(f\), so find the range first, and use the original domain to choose the sign of any square root.
Draw \(y = x\) lightly first, then reflect key points: intercepts \((0, c) \leftrightarrow (c, 0)\), end points \((a, b) \leftrightarrow (b, a)\), and swap the asymptotes. Any intersection of an increasing function with its inverse lies on \(y = x\).
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(\frac{x+6}{x+2}\) simplifies to 3. | 6 and 2 are terms, not factors of the whole numerator and denominator; at \(x = 1\) the fraction is \(\frac73\). | Cancel only common factors after factorising completely. |
| \(fg(x)\) means "do \(f\), then \(g\)". | \(fg(x) = f(g(x))\): \(g\) is applied to \(x\) first. | Work from the inside out: the function next to \(x\) acts first. |
| \(f^{-1}(x)\) is the same as \(\frac{1}{f(x)}\). | The inverse reverses the mapping; the reciprocal divides 1 by the output. For \(f(x) = 2x\), \(f^{-1}(x) = \frac{x}{2}\) but \(\frac{1}{f(x)} = \frac{1}{2x}\). | \(f^{-1}(x)\) undoes \(f\): \(f^{-1}f(x) = x\). |
| Every function has an inverse. | A many-one function would need its inverse to send one input to two outputs. | Only one-one functions have inverses; restrict the domain if necessary. |
| The domain of \(f^{-1}\) is the same as the domain of \(f\). | The roles of inputs and outputs are swapped. | Domain of \(f^{-1}\) = range of \(f\); range of \(f^{-1}\) = domain of \(f\). |
| The graph of \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in the \(x\)-axis or \(y\)-axis. | Reversing a mapping swaps the coordinates \((a, b) \to (b, a)\). | Reflect in the line \(y = x\). |
How this topic connects
Earlier pure units (P1, P2)
Factorising quadratics, completing the square and polynomial division by \((ax + b)\) are the tools used throughout this page; sketching curves with asymptotes (P1) is how ranges are found.
The rest of P3, and P4
The modulus function and transformations of graphs extend the function ideas here; \(e^x\) and \(\ln x\) are a function–inverse pair whose graphs reflect in \(y = x\); differentiating with the chain rule is differentiating a composite function; the quotient rule needs simplified algebraic fractions. In P4, partial fractions start from a proper fraction, so an improper fraction is first divided as above; the results feed binomial expansions and integration.
Mechanics and Physics
Models such as \(v = \frac{a}{t + b} + c\) in M2 and IAL Physics have a range set by an asymptote (a terminal value that is approached but never reached), found exactly as in Worked example 4.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c};\ \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}\) | dividing and adding algebraic fractions (use the lowest common denominator where possible) | No — learn it |
| \(\frac{N(x)}{D(x)} \equiv Q(x) + \frac{R(x)}{D(x)},\ \deg R < \deg D\) | dividing an improper fraction by a linear or quadratic denominator | No — learn it |
| \(fg(x) = f(g(x))\) | composite function: \(g\) first | No — learn it |
| \(f^{-1}f(x) = ff^{-1}(x) = x\) | defining and checking an inverse | No — learn it |
| domain of \(f^{-1}\) = range of \(f\); range of \(f^{-1}\) = domain of \(f\) | stating the domain and range of an inverse | No — learn it |
| \(y = f^{-1}(x)\) is the reflection of \(y = f(x)\) in \(y = x\) | sketching an inverse; swapping coordinates and asymptotes | No — learn it |
| \(ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}\) | range of a quadratic; choosing a restricted domain at the vertex | No — derive it |
Check your understanding
1. Simplify fully \(\dfrac{x^2 - 5x + 6}{x^2 - 9}\).
\(\frac{(x-2)(x-3)}{(x-3)(x+3)} = \frac{x-2}{x+3}\), for \(x \ne \pm 3\).
2. \(f(x) = 2x + 1\) and \(g(x) = x^2\), both with domain \(\mathbb{R}\). Find \(fg(x)\) and \(gf(x)\).
\(fg(x) = f(x^2) = 2x^2 + 1\); \(gf(x) = g(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1\). They are different.
3. Find the inverse of \(f(x) = \dfrac{5}{x - 3}\), \(x > 3\), and state its domain.
\(y = \frac{5}{x-3} \Rightarrow x - 3 = \frac5y \Rightarrow f^{-1}(x) = 3 + \frac5x\). The range of \(f\) is \(f(x) > 0\), so the domain of \(f^{-1}\) is \(x > 0\).
4. Explain why \(f(x) = x^2\), \(x \in \mathbb{R}\), has no inverse, and give a domain on which it does.
It is many-one: \(f(-2) = f(2) = 4\). Restricting to \(x \ge 0\) (or \(x \le 0\)) makes it one-one; then \(f^{-1}(x) = \sqrt{x}\) (or \(-\sqrt{x}\)), \(x \ge 0\).
5. Write \(\dfrac{x^2 + 3x + 5}{x + 1}\) in the form \(Ax + B + \dfrac{C}{x + 1}\).
\(x^2 + 3x + 5 = (x + 1)(x + 2) + 3\), so the fraction is \(x + 2 + \frac{3}{x+1}\): \(A = 1,\ B = 2,\ C = 3\).
6. \(h(x) = \dfrac{1}{x}\), \(x \ne 0\). Show that \(hh(x) = x\), and explain what this says about \(h^{-1}\).
\(hh(x) = h\left(\frac1x\right) = \frac{1}{1/x} = x\). Applying \(h\) twice returns the input, so \(h\) undoes itself: \(h^{-1}(x) = \frac1x = h(x)\). Its graph is symmetric in \(y = x\).
Specification checklist
- (P3 1.1) I can simplify an algebraic fraction by factorising fully and cancelling common factors, and I know which values of \(x\) are excluded.
- (P3 1.1) I can multiply, divide, add and subtract algebraic fractions whose denominators are linear or quadratic, giving the answer in its simplest form.
- (P3 1.1) I can use algebraic division (or comparing coefficients) to write an improper fraction as a polynomial plus a proper fraction, for linear and quadratic denominators.
- (P3 1.2) I can state the domain and find the range of a function, and recognise one-one and many-one mappings (and that a one-many mapping is not a function).
- (P3 1.2) I can form composite functions such as \(fg(x)\), knowing that \(g\) is applied first, and solve equations involving them.
- (P3 1.2) I can find an inverse function and its domain and range, sketch it as the reflection in \(y = x\), and restrict a domain so that an inverse exists.