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) composite Defining composites or open all working together.

(ii) inverse value

Inverse values

Hence or otherwise find (fg)−1(0)(fg)^{-1}(0).

Explore this idea

Hint

  1. (fg)−1(0)(fg)^{-1}(0) is the input xx with fg(x)=0fg(x)=0.

Answer · (ii) inverse value

(fg)−1(0)=e−1(fg)^{-1}(0)=e^{-1}.

Explore the idea

Open interactive exploration

Which input gives fg(x) = 0?

fg is strictly decreasing on (0, 1) from 1 to −1, so each height is hit once, and k = 0 is hit at x = e⁻¹.

Height of the horizontal linek = 0
Try an example
Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct intersections
1

Between −1 and 1 there is exactly one intersection. The function is one-one.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Pointxy
0.3678790
  • Fg
  • Y = k
  • Input x
  • Intersection selected in the coordinate table
Key idea

(fg)−1(0)(fg)^{-1}(0) means 'which input gives output 0?'. Reading off the graph or solving fg(x)=0fg(x)=0 gives the same answer, e−1e^{-1}.

Ideas to try
Set k=0k=0 and read the intersection: is it near e−1≈0.37e^{-1}\approx0.37?