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 (iii)(a) Recovering a function or open all working together.

(iii)(b)

Recovering a function

Find h(x)h(x) such that hg(x)=x2+1hg(x)=x^2+1.

Explore this idea

Hint

  1. hg(x)=x2+1hg(x)=x^2+1 gives hh evaluated at ln⁡x\ln x.

  2. Replace xx by g−1(x)=exg^{-1}(x)=e^x.

Answer · (iii)(b)

h(x)=e2x+1h(x)=e^{2x}+1, x<0x<0.

Values outside this interval are not fixed by the stated condition.

Explore the idea

Open interactive exploration

h(x) = eˣ and h(x) = e²ˣ + 1 on x < 0

Part (a) gives h = eˣ (from 0 up to 1). Part (b) gives h = e²ˣ + 1 (from 1 up to 2).

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • (a)
  • (b)
Ideas to try
Check: hg(x)hg(x) for (b) at x=12x=\frac12: h(ln⁡12)=e2ln⁡12+1=14+1h(\ln\frac12)=e^{2\ln\frac12}+1=\frac14+1, which equals x2+1x^2+1.