Recover the unknown function by undoing a logarithm

f:x\mapsto1+\frac2{x-1}

Total marks: 8

Solution

(i)

Existence of composites

$f(x)=1+\dfrac2{x-1}$ for $x<1$ and $g(x)=\ln x$ for $0<x<1$ . Explain why $gf$ does not exist.

Explore this idea

Hint

$gf$ exists only if $R_f\subseteq D_g$ .
Find $R_f$ from the graph: horizontal asymptote $y=1$ and vertical asymptote $x=1$ .

Answer · (i)

gf

Explore the idea

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

↓ f→ f →

f(x)

-1

↓ g→ g →

gf(x)

undefined

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

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

R_f=(-\infty,1)

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

Key idea

R_f\nsubseteq D_g

Ideas to try

Drag $x$ to $-1$ : what is $f(-1)$ ? Is it allowed?
For which $x$ is $f(x)$ allowed in $g$ ?

(ii) composite

Defining composites

Find an expression for $fg(x)$ .

Hint

Substitute $g(x)=\ln x$ into $f$ and combine the fractions.

Answer · (ii) composite

fg(x)=(\ln x+1)/(\ln x-1)

(ii) inverse value

Inverse values

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

Explore this idea

Hint

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

Answer · (ii) inverse value

(fg)^{-1}(0)=e^{-1}

Explore the idea

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

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.
Point	x	y
	0.367879	0

Fg
Y = k
Input x
Intersection selected in the coordinate table

Key idea

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

Ideas to try

Set

k=0

and read the intersection: is it near

e^{-1}\approx0.37

(iii)(a)

Recovering a function

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

Explore this idea

Hint

$gh=\text{identity}$ means $h$ undoes $g$ : $h=g^{-1}$ .

Answer · (iii)(a)

h(x)=e^x

Explore the idea

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

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<1

Range of f = domain of f⁻¹

y<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

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

Ideas to try

Why does

h

only exist for

x<0

(iii)(b)

Recovering a function

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

Explore this idea

Hint

$hg(x)=x^2+1$ gives $h$ evaluated at $\ln x$ .
Replace $x$ by $g^{-1}(x)=e^x$ .

Answer · (iii)(b)

h(x)=e^{2x}+1

Explore the idea

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.

Ideas to try

Check:

hg(x)

for (b) at

x=\frac12

h(\ln\frac12)=e^{2\ln\frac12}+1=\frac14+1

, which equals

x^2+1

Recover the unknown function by undoing a logarithm

Solution

Existence of composites

Range of $f$

Compare

f's outputs spill outside the domain of g

Defining composites

Substitute

Combine

Inverse values

Set the output to 0

Solve

Which input gives fg(x) = 0?

Recovering a function

Invert $g$

State $h$

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

Recovering a function

Put $x=g^{-1}(u)$

Simplify

Domain

h(x) = eˣ and h(x) = e²ˣ + 1 on x < 0

Range of fff

Compare

Substitute

Combine

Set the output to 0

Solve

Invert ggg

State hhh

Put x=g−1(u)x=g^{-1}(u)x=g−1(u)

Simplify

Domain

Range of $f$

Invert $g$

State $h$

Put $x=g^{-1}(u)$