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Identity rules still have domains · source reference

[2011/DHS/Prelim/I/4] · Question pp. 2 · Working pp. 10

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(a) If f is one-one, determine with justification whether ff−1ff^{-1} exists. [1] (b) g(x)=ln⁡xg(x)=\ln x for real x≥1x\ge1; h(x)=x2−2x+2h(x)=x^2-2x+2 for real x>0x>0. (i) Given gh exists, define it including its domain and range. [2] (ii) Sketch g, its inverse, and g−1gg^{-1}g together. [3] (iii) Find the inputs satisfying g−1g(x)=gg−1(x)g^{-1}g(x)=gg^{-1}(x). [1]

[2011/DHS/Prelim/I/4]

Printed working

Part (a)

ff is one-one, so f−1f^{-1} exists. Since Rf−1=Df⊆DfR_{f^{-1}}=D_f\subseteq D_f, the composite function ff−1ff^{-1} exists.

Part (b)(i)

Dgh=DhD_{gh}=D_h. Thus gh:x↦ln⁡(x2−2x+2)gh:x\mapsto\ln(x^2-2x+2), x>0x>0.

Rh=[1,∞)R_h=[1,\infty); the source sketch of hh labels minimum (1,1)(1,1) and an open point at (0,2)(0,2). The graph of gg starts at the closed point (1,0)(1,0).

Dh=(0,∞)→hRh=[1,∞)→gRgh=[0,∞).D_h=(0,\infty)\xrightarrow{h}R_h=[1,\infty)\xrightarrow{g}R_{gh}=[0,\infty).

Thus Rgh=[0,∞)R_{gh}=[0,\infty). The accompanying composition diagram places RhR_h inside DgD_g and arrows Dh→hRh→gRghD_h\xrightarrow{h}R_h\xrightarrow{g}R_{gh}, with an outer composite arrow ghgh.

Part (b)(ii)

On equal-scale axes, draw y=g(x)=ln⁡xy=g(x)=\ln x, x≥1x\ge1, starting at (1,0)(1,0); y=g−1(x)=exy=g^{-1}(x)=e^x, x≥0x\ge0, starting at (0,1)(0,1); and y=g−1g(x)=xy=g^{-1}g(x)=x, x≥1x\ge1, starting at (1,1)(1,1). The underlying reflection line y=xy=x is dashed where it is not part of the restricted composite.

Part (b)(iii)

g−1g(x)=gg−1(x)=x.g^{-1}g(x)=gg^{-1}(x)=x.

Thus x∈Dg∩Dg−1x\in D_g\cap D_{g^{-1}}, so x∈[1,∞)x\in[1,\infty).

Teacher margin notes

gh(x)=g(x2−2x+2)gh(x)=g(x^2-2x+2). To find RghR_{gh}: (1) find RhR_h using the graph of the first function hh; (2) sketch the second function gg; (3) RhR_h becomes the restricted domain of gg; (4) use RhR_h as the new domain and the graph of gg to find its range, which is RghR_{gh}.

When sketching ff and f−1f^{-1} on the same diagram, the scale must be the same for the x- and y-axes; their graphs must be symmetrical about y=xy=x. Always include endpoints. Dg−1g=Dg=[1,∞)D_{g^{-1}g}=D_g=[1,\infty).

Dgg−1=Dg−1=[0,∞)D_{gg^{-1}}=D_{g^{-1}}=[0,\infty). To compare the two composites, x∈Dg−1g∩Dgg−1⇒x∈Dg∩Dg−1x\in D_{g^{-1}g}\cap D_{gg^{-1}}\Rightarrow x\in D_g\cap D_{g^{-1}}.

Source diagrams

Original two range-finding graphs, quadratic h with minimum(1,1)/open(0,2), and restricted logarithm g with closed(1,0), including domain/range rays. Source composition diagram with Rh inside Dg, h andg arrows and compositegh. Restricted logarithm, exponential inverse and identity composite on equal-scale axes, with all three closed endpoints.

Provenance

Question page references: 2. Solution page references: 10. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.