Porisma
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Chapter exploration

Can one function feed another?

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.
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