Porisma

Why one order works and the other does not

The functions ff and gg are defined as

f(x)=1−x2f(x)=1-x^2, x∈Rx\in\mathbb R, −1≤x≤1-1\leq x\leq1,

g(x)=x+2g(x)=\sqrt{x}+2, x∈Rx\in\mathbb R, x≥0x\geq0.

(i) Show that the function gfgf exists.

(ii) Define the function gfgf and sketch the graph of gfgf, stating the range of gfgf.

(iii) Explain why fgfg does not exist.

CJC · Functions · Example 9 · 2025

After your attempt

Record how it went before checking the working. You can update this later.

Record confidence for 3 parts

(i) Show that gf exists

How did this part go?

(ii) Define, sketch and find the range

How did this part go?

(iii) Explain why fg does not exist

How did this part go?

Explore the graph

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.

Solution