Composite functions inquiry
The prompt
Mathematical inquiry processes: test particular cases; make conjectures about relationships; generalise and prove. Conceptual field of inquiry: function notation; inverse functions; composite functions.
The prompt was designed for a year 10 class as the basis for a short inquiry into composite functions. In showing two functions in general form and then an equation involving two composite functions, the prompt invites students to test particular cases.
As function notation is arbitrary knowledge (see a discussion of Hewitt's distinction between arbitrary and necessary knowledge here), the teacher might choose to introduce two preparatory concepts before expecting students to pose questions and make observations about the prompt.
(1) Evaluating functions
The lesson might start with the teacher contrasting an equation to a function (such as y = 3x - 2 and f(x) = 3x - 2) before going on to evaluate, for example, f(4) and f(10) for the same function. A more challenging approach would be to evaluate rational functions in the form:
In a structured inquiry, students could evaluate f(1), f(10), f(100), f(1000) and f(1 000 000) in order to find the limit as x tends to infinity (see the PowerPoint in 'Resources' below.) Would the limit be the same for f(-1), f(-10), f(-100) and so on?
(2) Inverse functions
The next stage is to introduce the class to the concept of an inverse function and to a procedure for finding one. If f(x) = y and g(y) = x, then g(y) is the inverse of f(x) and g(y) can be written as f-1(x). An example that leads into the form of the functions in the prompt is:
At this point, the teacher could prepare the ground for the composite functions in the prompt by emphasising that the relationship holds the other way round:
Later in the inquiry, the teacher could return to the example to show that inverse functions are a special case that satisfy the equation in the prompt:
November 2021
Sense-making, exploration and proof
Students start the inquiry by trying to make sense of the prompt. They draw on their existing knowledge to question or comment:
You could replace a, b, c, and d with numbers.
f(x) could equal 4x + 1 and g(x) could equal 3x - 2.
Does f(g(x)) mean you multiply the two functions?
Does f(g(x)) mean you put the functions together?
How do you combine two functions?
Is g(f(x)) the inverse of f(g(x))?
The teacher focuses on the three questions about the meaning of composite functions, explaining that f(g(x)) and g(f(x)) involve substituting one function into another. For f(g(x)), g(x) is the inner function and f(x) is the outer function. So, for the functions suggested in the orientation phase, we get
f(g(x)) = f(3x - 2) = 4(3x - 2) + 1 = 12x - 7 and
g(f(x)) = g(4x + 1) = 3(4x + 1) - 2 = 12x + 1
f(g(x)) = f(3x - 2) = 4(3x - 2) + 1 = 12x - 7 and
g(f(x)) = g(4x + 1) = 3(4x + 1) - 2 = 12x + 1
Students immediately notice that the coefficient of x is the same, but not the constant. Would this always be the case? The class decides to explore with the aim of creating a pair of functions for which f(g(x)) = g(f(x)). For students who require more structure, the teacher offers them pairs of functions to test.
The pair at the bottom of each column satisfies the condition in the prompt - that is, the composite functions are equal.
Conjectures and proof
Students start to develop conjectures from their examples. Some look at their examples, list those that 'work' and try to generalise. One student contends that if ad + b = bc + d, then the constants are equal. She presents her reasoning on the board:
f(g(x)) = f(cx + d) = a(cx + d) + b = acx + (ad + b) and
g(f(x)) = g(ax + b) = c(ax + b) + d = acx + (bc + d)
As we want f(g(x)) = g(f(x)), the constants must be equal and, therefore, ad + b = cb + d. She goes on to explain how she could use her formula to find values of a, b, c and d that form functions for which f(g(x)) = g(f(x)).
The teacher concludes the lesson by leading students in the co-construction of a formal proof:
Other lines of inquiry
1. Three functions
Are there three functions f(x) = ax + b, g(x) = cx + d and h(x) = ex + f such that f(g(h(x))) = h(g(f(x)))?
In this case the relationship between the variables simplifies to:
acf + ad + b = ecb + ed + f
2. Changing the degree of the functions
What would happen if one of the functions was quadratic, such as f(x) = ax + b and g(x) = cx2 + dx?
f(g(x)) = f(cx2 + dx) = a(cx2 + dx) + b = acx2 + adx + b
g(f(x)) = g(ax+ b) = c(ax + b)2 + d(ax + b) = a2cx2 + 2abcx + b2c + adx + bd
By matching terms in x2,
acx2 = a2cx2 which leads to a = 1
By matching terms in x and substituting in a = 1,
adx = 2abcx + adx
d = 2bc + d which leads to bc = 0 and b = 0
(c cannot equal zero, otherwise the function would not be quadratic).
So, in order to satisfy the equality, f(x) = x. In this case, c and d can take any value because f(g(x)) = g(f(x)) = cx2 + dx.
3. Inverse functions
Another line of inquiry involve inverse functions. The teacher might use the additional prompts to encourage students to verify particular cases and prove the general result. A proof of prompt 2 follows:
Prompt 1
Prompt 2
