Recover the unknown function by undoing a logarithm

Functions f:x↦1+2x−1 f:x\mapsto1+\frac2{x-1}, x∈Rx\in\mathbb R, x<1 x<1; g:x↦ln⁡xg:x\mapsto\ln x, x∈Rx\in\mathbb R, 0<x<10<x<1.

(i) Explain why composite gfgf does not exist. mark

(ii) Find an expression for fg(x) fg(x). Hence or otherwise find (fg)−1(0)(fg)^{-1}(0). marks

(iii) Find h(x) h(x) for each case: (a) gh(x)=xgh(x)=x, mark (b) hg(x)=x2+1 hg(x)=x^2+1. marks

[2018/DHS/Prelim/I/5]

Total marks: 8

Solution

If this part uses an earlier result, revisit (ii) inverse value Inverse values or open all working together.

(iii)(a)

Recovering a function

Find h(x)h(x) such that gh(x)=xgh(x)=x.

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Hint

  1. gh=identitygh=\text{identity} means hh undoes gg: h=g−1h=g^{-1}.

Answer · (iii)(a)

h(x)=exh(x)=e^x, x<0x<0.

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Open interactive exploration

ln x on (0, 1) and its inverse eˣ on x < 0

gh(x) = x means h is the mirror image of g in y = x.

Point on f: input xx = 0.40
Reset the controls

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))
(0.4, -0.91629)
Q = (f(x), x)
(-0.91629, 0.4)
on the inverse
Domain of f
0<x<10<x<1
Range of f = domain of f⁻¹
y<0y<0
Midpoint M on y = x
(-0.25815, -0.25815)
  • G(x) = ln x on (0, 1)
  • H(x) = eˣ for x < 0
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

h=g−1h=g^{-1}: domain and range swap, so hh lives on x<0x<0 and takes values in (0,1)(0,1).

Ideas to try
Why does hh only exist for x<0x<0?