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Worked solution

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

(i) Show that gf exists

(i) Rf=[0,1]⊆Dg=[0,∞)R_f=[0,1]\subseteq D_g=[0,\infty), so every output from ff can enter gg. Hence gfgf exists.

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(ii) Define, sketch and find the range

(ii) gf(x)=g(f(x))=1−x2+2gf(x)=g(f(x))=\sqrt{1-x^2}+2, for −1≤x≤1-1\leq x\leq1. Its graph is the upper semicircle of radius 1 centred at (0,2)(0,2), with closed endpoints (−1,2)(-1,2) and (1,2)(1,2) and maximum (0,3)(0,3). Thus Rgf=[2,3]R_{gf}=[2,3]. Equivalently, map Df=[−1,1]D_f=[-1,1] through ff to [0,1][0,1], then through gg to [2,3][2,3]. The linked lesson explorer traces this graph. For the mapping method, first find the range of f on its given domain, use that interval as the temporary input interval for g, then sketch g on that interval to read its outputs. This range calculation does not change the original definition of g.

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Four source graphs: f=1−x² on[−1,1] with closed endpoints (±1,0) and maximum (0,1); gf upper semicircle with endpoints (±1,2), maximum (0,3); g restricted temporarily to[0,1] from (0,2) to (1,3); full g starts at (0,2) and continues with range[2, infinity). Mapping gives[−1,1] to[0,1] to[2,3].
Four source graphs: f=1−x² on[−1,1] with closed endpoints (±1,0) and maximum (0,1); gf upper semicircle with endpoints (±1,2), maximum (0,3); g restricted temporarily to[0,1] from (0,2) to (1,3); full g starts at (0,2) and continues with range[2, infinity). Mapping gives[−1,1] to[0,1] to[2,3].
Four source graphs: f=1−x² on[−1,1] with closed endpoints (±1,0) and maximum (0,1); gf upper semicircle with endpoints (±1,2), maximum (0,3); g restricted temporarily to[0,1] from (0,2) to (1,3); full g starts at (0,2) and continues with range[2, infinity). Mapping gives[−1,1] to[0,1] to[2,3].
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(iii) Explain why fg does not exist

(iii) Rg=[2,∞)R_g=[2,\infty) is not a subset of Df=[−1,1]D_f=[-1,1]. Therefore fgfg does not exist on the given domain of gg.

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Common mistakes

Read gf from right to left: apply f first. Checking one successful input does not prove that every output from the first function is accepted by the second.