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Form a composite function and use it

  • In 5 of 20 recent exams
  • Paper 1 or 2
  • 2 to 3 marks
How it is set in the exam

Paper 1 (no calculator) mostly, 2–3 marks per part: "find (f∘g)(x)(f \circ g)(x)" as part (a), then a use of it. Typical (5 of 20 exam versions, one of them on Paper 2 (calculator), where the composite is then sketched on the GDC).

Spot it in the exam
Exam phrases to spot (4)

  • “find (f∘g)(x)(f \circ g)(x)”

  • “write down an expression for g(f(x))g(f(x))”

  • “show that (f∘g)(x)=…(f \circ g)(x) = \ldots”

  • “given that (g∘f)(3)=12(g \circ f)(3) = 12, find the possible values of kk”

In disguise (2)
  • the composite is asked for in its own part, only as the set-up for a later part: a quadratic in exe^x, a trig equation, or a sketch on the GDC. The later part is its own type's (Quadratic in disguise, Solve a trig equation on an interval, Sketch a function on a given domain with the GDC)
  • an exponential of a log, such as alog⁡ax+ca^{\log_a x + c}: the index laws turn it into a power of xx
Variants

the outer function a quadratic, cubic, trig, exponential or log function; the inner one linear, trig, exponential or log; a constant found from one value of the composite (a square, so two answers); f∘ff \circ f as an expression; a show-that; the composite then sketched on a given domain (: the sketch part is Sketch a function on a given domain with the GDC; Example 2.3 is that chain).

Looks like it, but isn't (7)
  • the composite is given as an expression for every xx, and the inner function (or its constants aa and bb) is asked for: Find the unknown function inside a composite
  • only a number is needed, such as (f∘g)(2)(f \circ g)(2) with nothing unknown: Values of f, f∘g and f⁻¹ from a graph or a formula
  • showing that f(f(x))=xf(f(x)) = x, or that f−1(x)=f(x)f^{-1}(x) = f(x) (the function is its own inverse), or finding a constant from it: Find the inverse function, with its domain
  • "g∘fg \circ f is a translation of ff", or describe the transformation from ff to g∘fg \circ f: Name the transformation: describe it or find its constants
  • the range of h(f(x))h(f(x)) when hh is linear, such as h(x)=x−5h(x) = x - 5 or h(x)=2x+1h(x) = 2x + 1: the graph of ff stretched or translated vertically, Transform a graph: new equation, image points, sketch, range
  • solving an equation built from a composite, when the solving needs another type's method (every angle on an interval, a quadratic in exe^x, an exponential equation): that type, whether or not an earlier part asked for the composite; the composite is its first line (Where composite and inverse functions hide)
  • differentiating a composite: Differentiate: choose the rule (the chain rule)
Worked example

A constant from one value of the composite: two answers

Paper 1 (no calculator)
Question

Let f(x)=x2−3f(x) = x^2 - 3 and g(x)=2x+kg(x) = 2x + k, where kk is a constant.
(a) Find (f∘g)(x)(f \circ g)(x), giving your answer in terms of kk. [2]
(b) Given that (f∘g)(1)=13(f \circ g)(1) = 13, find the possible values of kk. [3]
(c) For the positive value of kk, find (g∘f)(x)(g \circ f)(x). [2]

Try it yourself first, then compare.

Now try one like it

Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

A handwritten page as Paper Zero marks it: ticks at the two lines that earned a mark, a cross on line 3 where +7 was copied as −7, and a short note beside that line.123456f′(x) = 4x − 124x − 12 = 0, so x = 3f(3) = 2(3)² − 12(3) − 7= 18 − 36 − 7 = −25minimum at (3, −25)M1 ✓A1 ✓A1 ✗11The +7 became −7: 18 − 36 + 7 = −11.
Steps

  1. Which function is inside?

    In (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)) it is gg, the one next to xx.

  2. Replace every xx in the outer function by the inner expression, in brackets.

    Write this line before simplifying: it carries the method mark.

  3. Simplify as asked:

    expand; (ex)2=e2x(e^x)^2 = e^{2x}; alog⁡ax=xa^{\log_a x} = x; log⁡aax=x\log_a a^x = x; for a trig composite, a double angle identity from the formula booklet (SL 3.6).

  4. Use it:

    evaluate it, match a given form (a show-that, every step forwards), or set it equal to a value and solve (if the solving needs another type's method, that part belongs to that type: see Not this type). A square equal to a positive number gives two values: keep both unless a condition rules one out.

  5. On , a sketch:

    graph the composite on its domain (graph a function on a given domain) and follow the steps of Sketch a function on a given domain with the GDC, with the conventions in Sketching.

Common mistakes

  • The wrong order

    (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)), so gg goes into ff. Composing the other way round can keep a method mark, but every accuracy mark after it is lost.

  • No brackets

    with f(x)=x2−3f(x) = x^2 - 3 and g(x)=2x+kg(x) = 2x + k, f(g(x))=(2x+k)2−3f(g(x)) = (2x + k)^2 - 3, not 2x+k2−32x + k^2 - 3.

  • Expanding wrongly

    (2x+k)2=4x2+4kx+k2(2x + k)^2 = 4x^2 + 4kx + k^2, not 4x2+k24x^2 + k^2. The bracketed form (2x+k)2−3(2x + k)^2 - 3 already scores, so expand only if a form is asked for.

More mistakes (5)

  • Only one value of the constant

    (2+k)2=16(2 + k)^2 = 16 gives k=2k = 2 and k=−6k = -6.

  • (ex)2=ex2(e^x)^2 = e^{x^2}

    it is e2xe^{2x}.

  • Keeping an impossible value

    when a composite becomes a quadratic in exe^x: ex=−2e^x = -2 has no solution, because ex>0e^x > 0. Leaving it in costs the final .

  • Skipping steps in a show-that

    write each form, (23)log⁡2x\left(2^3\right)^{\log_2 x} and then (2log⁡2x)3\left(2^{\log_2 x}\right)^3, before the given result.

  • Degree mode

    for a trig composite on : check the settings.

Keep going

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