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Cambridge IGCSE0606

Functions

Mathematics Additional 0606 Chapter Notes

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Functions
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In this chapter

  1. 1Functions, one–one and many–one
  2. 2Function notation and composite functions
  3. 3Domain and range
  4. 4When does gf exist?
  5. 5Finding the inverse
  6. 6Domain and range of an inverse
  7. 7Graphs of a function and its inverse
  8. 8The modulus of a function

1. Functions, one–one and many–one

A function maps each value of x in the domain to one value of f(x). The domain is the set of inputs. The range (image set) is the set of outputs.

A one–one function has one input for each output. A many–one function has some output that comes from more than one input.

A many–one function has no inverse. To explain why something is a function, say that each x is mapped to a unique y. Having an input and an output is not enough.

Examiner insight

“The simplest explanations offered were based upon each x is mapped to a unique y, therefore a function, and the function being many-one, therefore no inverse.” “Some candidates considered that f was a function as it had input and output or a domain and range, misunderstanding what was required.”

Worked example 12 marks

The function f is defined by f(x) = x² for all real x.

(a) Explain why f is a function.

(b) Explain why f has no inverse.

  1. 1

    each value of x is mapped to one unique value of f(x) [1]

  2. 2

    f(2) = f(-2) = 4, so f is many–one, so it has no inverse [1]

2. Function notation and composite functions

fg(x) means f(g(x)). Apply g first, then f. The notation f²(x) means f(f(x)).

To form a composite function, replace x in the outer function by the whole inner function. It is not a product.

The order matters, so fg may not be the same as gf. Simplify at the end, and take care with powers: (e^(3x))² = e^(6x).

fg(x) = f(g(x))

f²(x) = f(f(x))

Examiner insight

“A few other candidates found the product of the functions, rather than the composite function, or composed the functions in the incorrect order.” “Some candidates used the incorrect order of operations.”

Worked example 13 marks

The function g is defined by g(x) = 2x² - 6x + 4 for x leqslant k, where k is a constant. The function h is defined by h(x) = e^(3x) - 1 for x leqslant 0.

Given that gh(x) exists, find gh(x), simplifying your answer.

  1. 1

    gh(x) = 2(e^(3x) - 1)² - 6(e^(3x) - 1) + 4 [1]

  2. 2

    = 2e^(6x) - 4e^(3x) + 2 - 6e^(3x) + 6 + 4 [1]

  3. 3

    = 2e^(6x) - 10e^(3x) + 12 [1]

3. Domain and range

The domain is the set of inputs, written as an inequality in x. The range is the set of outputs, written as an inequality in f(x) or y. Never give a range in x.

The domain may need to be restricted so that f⁻¹ exists. A quadratic is one–one on one side of its turning point, so complete the square to find that point and the least or greatest value.

For a composite function the domain of gf lies inside the domain of f, and the range of gf lies inside the range of g.

domain gf ⊆ domain f

range gf ⊆ range g

Examiner insight

“Those who did not gain full credit were sometimes confused about the correct notation for domain and range with the domain indicated as the range and vice versa.” “Some candidates made sign errors, notation errors or assumed the range was the set of real numbers.”

Worked example 14 marks

The function f is such that f(x) = 3x² + 8x for x geqslant k, where k is a constant. It is given that f⁻¹ exists. k has the least possible value for which f⁻¹ exists.

(a) Find the value of k.

(b) Find the range of f.

  1. 1

    3x² + 8x = 3(x + 4/3)² - 16/3 [1]

  2. 2

    turning point at x = -4/3, so k = -4/3 [1]

  3. 3

    least value of f(x) is -16/3 [1]

  4. 4

    range: f(x) geqslant -16/3 [1]

4. When does gf exist?

The composite gf exists when the range of f is a subset of the domain of g.

Apply f first. When gf exists, the domain of gf is the domain of f. It is not the domain of g.

To explain why gf exists, find the range of f and compare it with the domain of g. To find the least value of a constant, make the range of f fit fully inside the domain of g.

range f ⊆ domain g

Examiner insight

“The question depended on candidates realising that the range of f, a + 2, was fully included in the domain of g.” “Other candidates gave the domain of g rather than f as their answer.”

Worked example 14 marks

The function f is defined, for x geqslant -1, by f(x) = 1 + 2^(x). The function g is defined, for x geqslant -1, by g(x) = x(x +k) where k is a constant.

(i) Explain why gf exists.

(ii) Find an expression for gf(x).

(iii) It is given that gf(3) = 126. Find the value of k.

  1. 1

    f(x) geqslant 1.5, so the range of f is a subset of the domain of g [1]

  2. 2

    gf(x) = (1 + 2^(x))(1 + 2^(x) +k) [1]

  3. 3

    gf(3) = 9(9 +k) = 126 [1]

  4. 4

    9 + k = 14, so k = 5 [1]

5. Finding the inverse

Write y = f(x) and make x the subject. Then swap the variables, so the answer is in terms of x. Undo one operation at a time.

Use the correct notation: f⁻¹(x), not an expression in y. For a restricted quadratic, a square root gives ±. The domain of f shows which sign to keep.

f(x) = e^(2x) ⇒ f⁻¹(x) = 1/2ln x

Examiner insight

“Candidates who were the most successful, inverted the function using a series of steps, each of which changed a single operation.” “A few candidates omitted to swop the variables and offered an expression in y rather than x.”

Worked example 12 marks

The function g is defined by g(x) = 4e^(x-2) - 1 for all real x.

Find an expression for g⁻¹(x).

  1. 1

    y = 4e^(x-2) - 1, so e^(x-2) = (y + 1)/4 [1]

  2. 2

    x - 2 = ln(y+1)/4, so g⁻¹(x) = 2 + ln(x+1)/4 [1]

6. Domain and range of an inverse

An inverse undoes the function, so the domain of the inverse is the range of f, and the range of the inverse is the domain of f.

Write each in the right letter: a domain in x, a range in f⁻¹(x) or y.

domain f⁻¹ = range f

range f⁻¹ = domain f

Examiner insight

“The relationship between the domain and range of a function and its inverse was clearly not understood by many candidates.” “Many candidates did not give the range of the inverse function and those that did often used incorrect notation or had an incorrect answer.”

Worked example 14 marks

A function f is defined by f(x) = e^(2x+1) for x ∈ R.

(a) State the range of f⁻¹.

(b) Find the domain of f⁻¹.

(c) Find an expression for f⁻¹(x).

  1. 1

    range of f⁻¹ = domain of f, so f⁻¹(x) ∈ R [1]

  2. 2

    domain of f⁻¹ = range of f, so x > 0 [1]

  3. 3

    y = e^(2x+1), so ln y = 2x + 1 [1]

  4. 4

    f⁻¹(x) = (ln x - 1)/2 [1]

7. Graphs of a function and its inverse

The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x. Each function is the reflection of the other.

Sketch f first, then reflect it. Do not find the rule for the inverse unless the question asks. Reflect asymptotes and intercepts too, and give intercepts as exact values.

Examiner insight

“It was expected that candidates would reflect this curve in the line y = x rather than finding the rule for the inverse function which was not needed.” “A few candidates mentioned reflection but omitted to state the equation of the line or stated that the line was one of the axes or y = −x, for example.”

Worked example 14 marks

The function f is defined, for x geqslant -1, by f(x) = 1 + 2^(x).

Sketch the graph of y = f(x) and hence the graph of y = f⁻¹(x). State any intercepts with the coordinate axes.

  1. 1

    y = f(x): smooth rising curve starting at (-1, 1.5), drawn only for x geqslant -1 [1]

  2. 2

    intercept (0, 2) and no x-intercept [1]

  3. 3

    y = f⁻¹(x): reflection in the line y = x, starting at (1.5, -1) [1]

  4. 4

    intercept (2, 0) and no y-intercept [1]

8. The modulus of a function

The graph of y = |f(x)| is never below the x-axis. Where f(x) geqslant 0 it is the same as y = f(x). Where f(x) < 0 it is y = -f(x).

So the part of y = f(x) below the x-axis is reflected in the x-axis. Mark the intercepts clearly. Graphs of linear functions are ruled and graphs of quadratics are smooth.

Examiner insight

“As each of these graphs was the modulus of a linear function, both graphs should have been ruled.” “Some candidates drew their local maximum on the y-axis.”

Worked example 14 marks

Sketch the graph of y = |x² - 2x - 3|.

State the intercepts with the coordinate axes.

  1. 1

    x² - 2x - 3 = (x - 3)(x + 1) = 0, so x-intercepts (-1, 0) and (3, 0) [1]

  2. 2

    y-intercept: |-3| = 3, so (0, 3) [1]

  3. 3

    between -1 and 3 the curve is reflected in the x-axis, with maximum point (1, 4) [1]

  4. 4

    smooth curve, as an upside-down U between -1 and 3 and rising outside [1]

Recap

  • A function maps each x to one unique y. Many–one means no inverse.
  • fg(x) = f(g(x)): apply g first. f²(x) = f(f(x)). It is not a product.
  • Domain in x. Range in f(x) or y. Restrict a quadratic at its turning point.
  • gf exists when range of f is inside domain of g. Then domain of gf = domain of f.
  • Inverse: make x the subject, then swap. Give the answer in x.
  • Domain of f⁻¹ = range of f. Range of f⁻¹ = domain of f.
  • Graph of f⁻¹ = reflection of f in y = x.
  • y = |f(x)|: the part below the x-axis is reflected in the x-axis.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. The function f is defined by f(x) = 3x - 1 for x ∈ {1, 2, 3, 4}. Write down the range of f.2 marks
  2. The function f is defined by f(x) = (x - 2)² for all real x. Explain why f has no inverse.2 marks
  3. The functions f and g are defined by f(x) = 2x + 3 and g(x) = x² for all real x. Find (a) fg(x), (b) gf(x), (c) f²(x).3 marks
  4. Do not use a calculator. The function f is defined by f(x) = x² - 6x + 5 for all real x. Find the range of f.3 marks
  5. Do not use a calculator. For the graph of y = |x² - 4|: (a) write down the x-intercepts, (b) write down the y-intercept of y = x² - 4 and of y = |x² - 4|, (c) describe what happens to the part of y = x² - 4 below the x-axis.3 marks
  6. Do not use a calculator. The function f is defined by f(x) = 2e^(x) + 3 for all real x. (a) Find f⁻¹(x). (b) State the geometrical relationship between y = f(x) and y = f⁻¹(x).4 marks
  7. Do not use a calculator. The function f is defined by f(x) = (x + 4)² - 1 for x leqslant -4. Find f⁻¹(x) and state its domain.4 marks
  8. Do not use a calculator. The function f is defined by f(x) = x² + 2 for all real x. The function g is defined by g(x) = √(x - 1) for x geqslant 1. (a) Explain why gf exists. (b) Find gf(x). (c) Find fg(x), simplified, and state its domain.4 marks
  9. Do not use a calculator. A toy rocket is fired upwards. Its height above the ground, in metres, t seconds after firing is f(t) = 20 - (t - 4)² for 0 leqslant t leqslant 4. (a) Explain why f would have no inverse if its domain were 0 leqslant t leqslant 8. (b) For the domain 0 leqslant t leqslant 4, find f⁻¹(x) and state its domain and its range. (c) Use f⁻¹ to find the time at which the height is 11 m.8 marks
  10. The function f is defined by f(x) = e^(2x) + a for x geqslant 0, where a is a constant. The function g is defined by g(x) = x² - 6x for x geqslant 3. (a) Find the range of f in terms of a. (b) Find the least value of a for which gf exists. For this value of a: (c) find gf(x), simplifying your answer, (d) state the domain of gf and find the range of gf, (e) solve gf(x) = 7, giving your answer in exact form.10 marks

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