Porisma

Can one function feed another?

Follow the order of a composite, check every intermediate output, and map the final range.

← Back to lesson · Complete chapter reading and print view

By the end, you can

  • Read composition from right to left
  • Check whether the first range fits inside the second domain
  • Find the composite domain and range

about 20 minutes for the lesson · practise at your own pace.

The central idea

In gf(x)=g(f(x))gf(x)=g(f(x)), apply ff first, then pass its output to gg. This works on the entire original domain of ff only when Rf⊆DgR_f\subseteq D_g. If even one output is outside the next function’s domain, the composite does not exist on that whole domain. A restriction can define a different composite, but must be stated explicitly.

Composition maps into reached subsets

Expand diagram

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x in Dg=Dfg passes through g into a subset of Df, then f into a subset of Rf. The direct fg arrow reaches the same output; full outer sets remain distinct from reached subsets.
x in Dg=Dfg passes through g into a subset of Df, then f into a subset of Rf. The direct fg arrow reaches the same output; full outer sets remain distinct from reached subsets.
x in Dg=Dfg passes through g into a subset of Df, then f into a subset of Rf. The direct fg arrow reaches the same output; full outer sets remain distinct from reached subsets.
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Explore

Here f takes x to 1 − x², then g takes that intermediate output to its square root plus 2. Move x from one endpoint to the other and watch the intermediate value and final point.

x → f(x) → g(f(x))

Trace Example 9 without changing its printed domain.

Input x (in the domain of f)x = 0.00
Try an example
input
0
↓ f
f(x)
1
↓ g
gf(x)
3

1 is an allowed input of g.

D_gR_f-11.54

D_g: [0, ∞); R_f: [0, 1]. Finite bounds are shown to 6 significant figures. Hollow ends are excluded; arrows continue beyond this window.

Selected output: 1.

The composite exists

Rf=[0,1]⊆Dg=[0,∞)R_f=[0,1]\subseteq D_g=[0,\infty); Dgf=[−1,1]D_{gf}=[-1,1] and Rgf=[2,3]R_{gf}=[2,3].

Key idea

Every output from f lies in the domain of g, so gf exists on all of [-1,1].

Ideas to try
  1. Compare the two endpoint inputs: both reach final output 2.
  2. At x = 0, the intermediate value is 1 and the final value is 3.

Worked example

For f(x)=1−x2f(x)=1-x^2 on [−1,1][-1,1], Rf=[0,1]R_f=[0,1]. With g(t)=t+2g(t)=\sqrt{t}+2 on [0,∞)[0,\infty), every intermediate value is allowed. Thus gf(x)=1−x2+2gf(x)=\sqrt{1-x^2}+2, with domain [−1,1][-1,1] and range [2,3][2,3]. In the opposite order, Rg=[2,∞)R_g=[2,\infty) does not fit inside Df=[−1,1]D_f=[-1,1], so fgfg does not exist on the given domain of gg.

Substitution gives a candidate formula; it does not establish that the composite exists on the printed domain.

Check your understanding

Let a(x)=x2a(x)=x^2 on R\mathbb R and b(t)=t−1b(t)=\sqrt{t-1} on [1,∞)[1,\infty). Decide whether baba and abab exist on their stated domains. Justify each decision using the entire inner range, and give the rule, domain and range of any composite that exists.

Practice questions

Practice section 1 · Questions 1–6

Follow the order and the reached range · Identity rules can have different domains · +4 more

Practice section 2 · Questions 7–9

A piecewise inverse and two identity domains · Check whole-domain composition before finding its range · +1 more

More practice

Use the range/domain check on these questions, then work through their existing interactive solutions.

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