1. Function notation
Inputs and outputs
- A function is a rule that takes a number in and gives exactly one number back
- The number going in is the input and the number coming back is the output
- Writing f(x) = 4x − 7 names the rule f and says what it does to whatever is put in
- Any letter can name a function, and a question that needs a second rule simply uses a second letter such as g or h
- Every x in the rule is replaced by whatever sits inside the brackets, whether that is a number, a letter or a whole expression
- f(3) = 4 × 3 − 7 = 5
- f(−2) = 4 × (−2) − 7 = −15
- f(a) = 4a − 7, which cannot be tidied any further
- Bracket a negative input as you substitute it, otherwise the sign is easily dropped
Reading the question the right way round
- f(6) means put 6 in and see what comes out
- f(x) = 6 means the output is 6 and the input is unknown, so it is an equation to solve
- These two look almost identical on the page and mean opposite things, so read which side the number is on
- To solve f(x) = 21 for f(x) = 4x − 7, replace f(x) by the rule: 4x − 7 = 21, so 4x = 28 and x = 7
- A mapping diagram shows the same idea as two columns joined by arrows, inputs on the left and outputs on the right, with one arrow leaving each input
- For f(x) = x + 3 the arrows run 3 → 6, 4 → 7 and, in general, x → x + 3
Worked example
Substituting into a function, and solving one
f(x) = 3x − 5. Find f(4), and solve f(x) = 16.
Solution:
- f(4) means substitute 4 wherever x appears
- f(4) = 3 × 4 − 5 = 12 − 5 = 7
- f(x) = 16 is a different demand: it is an equation to solve, not a substitution
- 3x − 5 = 16
- Add 5 to both sides: 3x = 21
- Divide by 3: x = 7
- The two answers are both 7 here by coincidence, which is exactly why the difference has to be read carefully: f(4) asks for an output, f(x) = 16 asks for an input
2. Domain and range
What each word covers
- The domain is the collection of inputs a function is allowed to take
- The range is the collection of outputs those inputs produce
- A domain is always described in terms of x, and a range always in terms of f(x), so keep the two letters straight
- On this specification the domain is very often handed to you as a short list of numbers inside curly brackets, and the range is then simply the list of answers you get by substituting each one
- The outputs are listed in the same curly-bracket form, and getting most of them right still scores
Worked example
A range from a listed domain
g(x) = x² + 2 has domain {−3, 1, 4}. Find the range of g(x).
Solution:
- The domain is the set of inputs allowed, and the range is the set of outputs they produce
- The domain here is a short list, so the range is a short list too: substitute each value in turn
- g(−3) = (−3)² + 2: bracket the negative, then square it
- (−3)² = 9, so g(−3) = 9 + 2 = 11
- g(1) = 1² + 2 = 1 + 2 = 3
- g(4) = 4² + 2 = 16 + 2 = 18
- The range is {3, 11, 18}
- Give the outputs, not the inputs: {−3, 1, 4} is the domain and answers a different question
- Writing them in increasing order is conventional, and it also makes a missing value obvious
Domains given as an interval, and values that must be excluded
- Where the domain is an interval such as 1 ⩽ x < 5, substitute each end value to find where the outputs start and stop, provided the graph only rises or only falls across the interval (when a quadratic's turning point lies inside the interval, the turning-point value is one end of the range and the end value further from it is the other)
- Carry the ⩽ and < across carefully, remembering that a decreasing function swaps which end is included
- Some values can never be inputs, and the two reasons are dividing by zero and square rooting a negative
- h(x) = 5 ÷ (x − 6) excludes x = 6, because the bottom would be zero
- j(x) = √(x − 2) excludes everything below 2, because the inside would be negative
- To find an excluded value, set the bottom of the fraction equal to zero and solve
- A quick sketch of the graph settles most range questions faster than substituting does
- Composite functions are the one exception here: this specification does not ask for their domains or ranges
Worked example
A range from an interval
f(x) = 4x − 7 has domain 1 ⩽ x < 5. Find the range of f(x).
Solution:
- With an interval there are too many inputs to list, so work from the ends
- The rule multiplies by 4, which is positive, so larger inputs give larger outputs and the order of the ends is preserved
- At the lower end: f(1) = 4 − 7 = −3
- At the upper end: f(5) = 20 − 7 = 13
- Each end keeps the symbol it had: 1 was included, so −3 is included; 5 was excluded, so 13 is excluded
- The range is −3 ⩽ f(x) < 13
- Write the range in terms of f(x), not x: −3 ⩽ x < 13 describes a set of inputs
- Had the rule multiplied by a negative number the ends would swap over, so the check is always to substitute both and see which output is larger
3. Composite functions
Reading the order
- A composite function feeds the output of one function straight into another
- gf(x) is shorthand for g(f(x)), and the function written next to the bracket acts first
- So gf means f then g, and fg means g then f, which is the reverse of the order most people read it in
- Rewriting gf(x) as g(f(x)) before doing anything is the surest way to avoid reversing it
- ff(x) applies the same function twice, and it is not the same as squaring the output
Working one out
- With a number, work out the inner function first and put that answer into the outer one
- With algebra, substitute the whole inner expression into the outer rule and then expand
- The substitution line scores a mark before any expanding is done, so write it out
Worked example
Numbers and algebra
f(x) = 4x − 7 and g(x) = x² + 2. Find gf(2), fg(2) and gf(x).
Solution:
- For gf(2), f acts first: f(2) = 8 − 7 = 1
- Then g acts on that: g(1) = 1 + 2 = 3, so gf(2) = 3
- For fg(2), g acts first: g(2) = 4 + 2 = 6
- Then f acts on that: f(6) = 24 − 7 = 17, so fg(2) = 17
- For gf(x), substitute the whole of f(x) into g: g(4x − 7) = (4x − 7)² + 2
- Expand the bracket: 16x² − 56x + 49 + 2
- So gf(x) = 16x² − 56x + 51
Worked example
Solving a composite equation
f(x) = 4x − 7 and g(x) = x² + 2. Solve fg(x) = 37.
Solution:
- fg(x) = f(g(x)), so g acts first and the whole of its output goes into f
- fg(x) = 4(x² + 2) − 7
- Expand and collect: 4x² + 8 − 7 = 4x² + 1
- Now solve 4x² + 1 = 37
- Subtract 1: 4x² = 36, then divide by 4: x² = 9
- Square root both sides, keeping both signs: x = 3 or x = −3
- Check: g(3) = 11 and f(11) = 44 − 7 = 37, and g(−3) is also 11
- Undoing f first gives the same equation: 4 × g(x) − 7 = 37, so g(x) = 11 and x² + 2 = 11
- If the question asks only for the positive solution, give x = 3 alone
4. Inverse functions
Undoing the rule
- The inverse of f, written f⁻¹, reverses what f does, so it turns each output back into the input that produced it
- An output that came from a given input goes straight back to it, so f(a) = b guarantees f⁻¹(b) = a, a check that needs no algebra
- Applying a function and then its inverse leaves the number untouched: ff⁻¹(x) = x and f⁻¹f(x) = x
- That cancelling is the fastest route through a question asking to solve f⁻¹(x) = c, since applying f to both sides gives x = f(c) directly
- The −1 here is not a power, so f⁻¹(x) is not 1 ÷ f(x)
Worked example
Solving h⁻¹(x) = c without finding the inverse
h(x) = 3ˣ. Find x when h⁻¹(x) = 4, and find x when h⁻¹(x) = −2.
Solution:
- h⁻¹(x) = 4 says the inverse sends x to 4, so h sends 4 back to x
- So x = h(4): write this line first, because it earns a method mark on its own
- x = 3⁴ = 81
- In the same way, h⁻¹(x) = −2 gives x = h(−2) = 3⁻² = 1 ÷ 3² = 1/9
- Check: h(4) = 81, so h⁻¹ takes 81 back to 4, as the question said
- There is no need to find h⁻¹(x), and for 3ˣ you could not: swapping the letters gives x = 3ʸ, which no rearranging in this course can solve for y
Finding it algebraically
- Replace f(x) with y
- Swap every x and every y, which changes the letters without moving any term
- Rearrange until y stands alone, using exactly the methods for changing the subject of a formula
- Write the answer as f⁻¹(x) = …, with no y left anywhere
- The first rearranging step earns a mark by itself, so show it
Worked example
Finding an inverse and checking it
f(x) = 4x − 7. Find f⁻¹(x).
Solution:
- The inverse undoes the function, so it turns an output back into the input that produced it
- Write the rule as an equation: y = 4x − 7
- Swap the letters, because the inverse reverses the roles of input and output: x = 4y − 7
- That swapped line scores a mark on its own, so put it down before rearranging
- Now make y the subject. Add 7 to both sides: x + 7 = 4y
- Divide both sides by 4, and the bracket matters because the whole of x + 7 is divided
- y = (x + 7) ÷ 4, so f⁻¹(x) = (x + 7) ÷ 4
- Check by running a value through both: f(3) = 12 − 7 = 5, and f⁻¹(5) = 12 ÷ 4 = 3, which is the input we started from
- Note f⁻¹ means the inverse function, never 1 ÷ f(x)
Domain and range swap over
- The inverse sends outputs back to inputs, so the two sets trade places
- The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f
- Rewrite the swapped set in the right letters: a domain in terms of x, a range in terms of f⁻¹(x)
- With f(x) = 4x − 7 on 1 ⩽ x < 5 the range is −3 ⩽ f(x) < 13, so f⁻¹ has domain −3 ⩽ x < 13