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

(i)

Existence of composites

f(x)=1+2x−1f(x)=1+\dfrac2{x-1} for x<1x<1 and g(x)=ln⁡xg(x)=\ln x for 0<x<10<x<1. Explain why gfgf does not exist.

Explore this idea

Hint

  1. gfgf exists only if Rf⊆DgR_f\subseteq D_g.

  2. Find RfR_f from the graph: horizontal asymptote y=1y=1 and vertical asymptote x=1x=1.

Answer · (i)

gfgf does not exist: Rf=(−∞,1)⊈(0,1)=DgR_f=(-\infty,1)\nsubseteq(0,1)=D_g.

Explore the idea

Open interactive exploration

f's outputs spill outside the domain of g

Drag x. Whenever f(x) ≤ 0, it is not an allowed input of g = ln on (0, 1).

Input x (in the domain of f)x = 0.00
Try an example
input
0
↓ f
f(x)
-1
↓ g
gf(x)
undefined

-1 is outside the domain of g, so gf(x) does not exist for this x.

D_gR_f-4-12

D_g: (0, 1); R_f: (−∞, 1). Finite bounds are shown to 6 significant figures. Hollow ends are excluded; arrows continue beyond this window.

Selected output: -1.

The composite does not exist as defined

Rf=(−∞,1)R_f=(-\infty,1) is not inside Dg=(0,1)D_g=(0,1): all the negative outputs are the problem.

  • F(x)
  • Gf(x), where it exists
  • Lower end of D_g
Key idea

A single bad output is enough: Rf⊈DgR_f\nsubseteq D_g, so gfgf does not exist.

Ideas to try
  1. Drag xx to −1-1: what is f(−1)f(-1)? Is it allowed?
  2. For which xx is f(x)f(x) allowed in gg?