Identity rules still have domains

(a) If f is one-one, determine with justification whether ff−1ff^{-1} exists. mark

(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. marks

(ii) Sketch g, its inverse, and g−1gg^{-1}g together. marks

(iii) Find the inputs satisfying g−1g(x)=gg−1(x)g^{-1}g(x)=gg^{-1}(x). mark

[2011/DHS/Prelim/I/4]

Total marks: 7

Solution

(a)

Existence of composites

If ff is one-one, determine with justification whether ff−1ff^{-1} exists.

Hint

  1. ff−1ff^{-1} means apply f−1f^{-1} first, then ff.

  2. It exists if Rf−1⊆DfR_{f^{-1}}\subseteq D_f.

Answer · (a)

Yes: Rf−1=DfR_{f^{-1}}=D_f, so ff−1ff^{-1} exists.

(b)i

Defining a composite

g(x)=ln⁡xg(x)=\ln x for real x≥1x\ge1 and h(x)=x2−2x+2h(x)=x^2-2x+2 for real x>0x>0. Given that ghgh exists, define ghgh, including its domain and range.

Explore this idea

Hint

  1. The rule: substitute hh into gg.

  2. The domain of ghgh is the domain of the first function, hh.

  3. The range: feed RhR_h into gg.

Answer · (b)i

gh(x)=ln⁡(x2−2x+2)gh(x)=\ln(x^2-2x+2), x>0x>0; range [0,∞)[0,\infty).

Explore the idea

The outputs of h land in the domain of g

Drag x. R_h = [1, ∞) fits exactly inside D_g = [1, ∞).

Input x (in the domain of h)x = 1.00
Try an example
input
1
↓ h
h(x)
1
↓ g
gh(x)
0

1 is an allowed input of g.

D_gR_h0612

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

Selected output: 1.

The composite exists

Rh=[1,∞)⊆Dg=[1,∞)R_h=[1,\infty)\subseteq D_g=[1,\infty): the smallest output of hh is exactly the first allowed input of gg, and the composite has range [0,∞)[0,\infty).

  • H(x)
  • Gh(x)
  • Lower end of D_g
Key idea

The composite is defined because hh never outputs anything below 1, which is the smallest input gg accepts.

Ideas to try
Drag xx to 1: what is h(1)h(1) and gh(1)gh(1)?
(b)ii

Graphs of inverses

Sketch y=g(x)y=g(x), y=g−1(x)y=g^{-1}(x) and y=g−1g(x)y=g^{-1}g(x) on the same diagram, with the same scale on both axes.

Explore this idea

Hint

  1. g=ln⁡xg=\ln x for x≥1x\ge1 starts at (1,0)(1,0). Its inverse is exe^x with domain x≥0x\ge0.

  2. g−1g(x)=xg^{-1}g(x)=x with domain x≥1x\ge1.

  3. Mark each starting point.

g,g⁻¹ andg⁻¹g on equal-unit axes
g,g⁻¹ andg⁻¹g on equal-unit axesScroll across to see the full diagram.

Answer · (b)ii

g=ln⁡xg=\ln x starts (1,0)(1,0); g−1=exg^{-1}=e^x starts (0,1)(0,1); g−1g=xg^{-1}g=x starts (1,1)(1,1).

Domains x≥1, x≥0, x≥1 respectively.

Explore the idea

ln x and eˣ reflected in y = x

Drag the input. The composite g⁻¹g is the part of the mirror line y = x that starts at (1, 1).

Point on f: input xx = 2.00
Try an example

Reflection swaps the coordinates. The segment PQ is perpendicular to y = x, and its blue midpoint M lies on that line.

P = (x, f(x))
(2, 0.69315)
Q = (f(x), x)
(0.69315, 2)
on the inverse
Domain of f
x≥1x\ge1
Range of f = domain of f⁻¹
y≥0y\ge0
Midpoint M on y = x
(1.34657, 1.34657)
  • G(x) = ln x, x ≥ 1
  • G⁻¹(x) = eˣ, x ≥ 0
  • Y = x, which contains the composite g⁻¹g for x ≥ 1
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

Domains swap: gg starts at x=1x=1, so g−1g^{-1} starts at y=1y=1; gg's range starts at 0, so g−1g^{-1}'s domain starts at 0.

Ideas to try
Where does g−1g^{-1} start? Why not at (1,e)(1,e)?
(b)iii

Equal composites

Find the values of xx for which g−1g(x)=gg−1(x)g^{-1}g(x)=gg^{-1}(x).

Hint

  1. Each composite is xx on its own domain.

  2. The equality holds where both are defined.

Answer · (b)iii

x≥1x\ge1.