Explainer: Composite Functions

In this explainer, we will learn how to form a composite function by composing two or more linear, quadratic, exponential, or radical functions.

A cycling tour takes you along a mountainous path. Your position after 𝑑 minutes of cycling is given by the expression 𝑃(𝑑). Suppose that what we are really interested in terms of β€œposition” is just your altitude, say 𝑧, in meters above sea level. So, 𝑧=𝑃(𝑑) gives your altitude 𝑑 minutes into your cycle.

Suppose, also, that in the region where you are cycling, the temperature (in degrees Celsius) depends only on altitude. So, no matter where exactly you are, your altitude determines your temperature. In other words, the temperature 𝑇(𝑧) is a function of altitude 𝑧.

If, after 12 minutes, you are at altitude 335 m and it is know that the temperature at 335 m is 14∘C, it follows that the temperature you experience after 12 minutes is 14∘C.

We can β€œcompose” these two functions to give the temperature around you at time 𝑑 by π‘‘βŸΆπ‘ƒ(𝑑)=π‘§βŸΆπ‘‡(𝑧)=𝑇(𝑃(𝑑)) that gives us the new function π‘‘βŸΆπ‘‡(𝑃(𝑑)), which is given the β€œname” π‘‡βˆ˜π‘ƒ that is read as β€œπ‘‡ of 𝑃”.

In terms of expressions, if 𝑔(π‘₯)=3π‘₯+πœ‹ while 𝑓(π‘₯)=2π‘₯+π‘₯βˆ’1, then π‘“βˆ˜π‘” is produced by β€œsubstituting” every occurrence of π‘₯ in 𝑓(π‘₯) with 𝑔(π‘₯): (π‘“βˆ˜π‘”)(π‘₯)=𝑓(𝑔(π‘₯))=𝑓(3π‘₯+πœ‹)=2(3π‘₯+πœ‹)+(3π‘₯+πœ‹)βˆ’1=2ο€Ή9π‘₯+6πœ‹π‘₯+πœ‹ο…+3π‘₯+πœ‹βˆ’1=18π‘₯+(12πœ‹+3)π‘₯+ο€Ή2πœ‹+πœ‹βˆ’1.

Example 1: Evaluating Composite Functions at a Given Value

Given that 𝑓(π‘₯)=3π‘₯βˆ’1 and 𝑔(π‘₯)=π‘₯+1, find (π‘“βˆ˜π‘”)(2).


By definition, (π‘“βˆ˜π‘”)(π‘₯)=𝑓(𝑔(π‘₯)). So, here (π‘“βˆ˜π‘”)(2)=𝑓(𝑔(2))=𝑓2+1𝑔(π‘₯))=𝑓(5)=3(5)βˆ’1𝑓(π‘₯))=14.(bythedescriptionof(bythedescriptionof

So, (π‘“βˆ˜π‘”)(2)=14.

Note that we could have worked out the expression and found (π‘“βˆ˜π‘”)(π‘₯)=3π‘₯+2 and then gotten (π‘“βˆ˜π‘”)(2)=3(2)+2=14. This, however, does more than is asked of us.

The next example is not very different in principle.

Example 2: Evaluating Composite Functions at a Given Value

Set 𝑓=π‘“βˆ˜π‘“,𝑓=π‘“βˆ˜π‘“βˆ˜π‘“()(), and so on, so that 𝑓(π‘₯)=𝑓(𝑓(π‘₯))() and 𝑓(π‘₯)=𝑓(𝑓(𝑓(π‘₯)))(), and so on. Suppose that 𝑔(π‘₯)=4π‘₯βˆ’5. Find 𝑔(3)(οŠͺ).


Finding the expression for 𝑔(π‘₯)(οŠͺ) first looks like hard work. We will instead β€œfollow” the values: 𝑔(3)=4(3)βˆ’5=7βŸΉπ‘”(3)=𝑔(𝑔(3))=𝑔(7)=4(7)βˆ’5=23βŸΉπ‘”(3)=𝑔(23)=4(23)βˆ’5=87βŸΉπ‘”(3)=𝑔(87)=4(87)βˆ’5=343.()()(οŠͺ)

Observe that there are different ways of representing functions. One of them is by ordered pairs. For example, consider the two functions 𝑓 and 𝑔 given by 𝑓={(1,π‘Ž),(2,𝑏),(3,𝑏),(4,𝑐)} and 𝑔={(π‘Ž,4),(𝑏,3),(𝑐,2),(𝑑,1)}.

We can determine each of

  1. (π‘“βˆ˜π‘”)(π‘Ž),
  2. (π‘”βˆ˜π‘“)(3),
  3. π‘”βˆ˜π‘“.

Indeed, (π‘“βˆ˜π‘”)(π‘Ž)=𝑓(𝑔(π‘Ž))=𝑓(4)=𝑐,

and (π‘”βˆ˜π‘“)(3)=𝑔(𝑓(3))=𝑔(𝑏)=3.

And for π‘”βˆ˜π‘“, we will return another list of ordered pairs. The domain of π‘”βˆ˜π‘“ is the same as the domain of 𝑓, so {1,2,3,4}. The range will be a subset of the range of 𝑔, which is the same set.

From the above, we found that (π‘”βˆ˜π‘“)(3)=3, so (3,3) is one of these pairs. We complete this to π‘”βˆ˜π‘“={(1,4),(2,3),(3,3),(4,2)}.

Another way to represent functions is graphically.

Example 3: Evaluating Composite Functions at a Given Value From Their Graphs

In the figure, the red graph represents 𝑦=𝑓(π‘₯) while the blue graph represents 𝑦=𝑔(π‘₯).

What is 𝑓(𝑔(2))?


From the blue graph, we see that the point with π‘₯=2 is (2,1), so 𝑔(2)=1. From the red graph, we see that the point with π‘₯=1 is (1,3), so 𝑓(1)=3.

Therefore, 𝑓(𝑔(2))=3.

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