Matching rules are not equal without matching domains

f

Total marks: 11

Solution

(i)

One-one restrictions

$f(x)=e^{|2-x|}$ for real $x>a$ . Given that $a$ is the least value such that $f^{-1}$ exists, state $a$ .

Explore this idea

Hint

The graph of $e^{|2-x|}$ is V-shaped: falling to a minimum at $x=2$ , then rising.
A one-one piece must stay on one side of the minimum.

Answer · (i)

a=2

Explore the idea

Where does the graph stop being one-one?

Above the minimum height 1, a horizontal line cuts the full graph twice, once on each side of x = 2.

Height of the horizontal linek = 4

Try an example

Test boundary heights

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

Number of distinct intersections

Above 1 the full graph has two intersections, one on each side of x = 2.

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
	0.613706	4
	3.38629	4

Y = e^|2−x|
Y = k
Intersections
Intersection selected in the coordinate table

Key idea

Every height above the minimum is hit twice. Starting the domain at $x=2$ removes the left arm.

Ideas to try

Find the height at which there is exactly one intersection.

(ii)

Inverse functions

Find $f^{-1}(x)$ and state its domain.

Hint

For $x>2$ , $|2-x|=x-2$ .
Solve $y=e^{x-2}$ for $x$ .
$D_{f^{-1}}=R_f$ .

Answer · (ii)

f^{-1}(x)=\ln x+2

(iii)

Graphs of inverses and composites

Sketch $y=f(x)$ , $y=f^{-1}(x)$ and $y=ff^{-1}(x)$ on the same diagram, giving the coordinates of all end-points.

Explore this idea

Hint

$f$ : starts just right of $(2,1)$ and rises. $f^{-1}$ : reflection, starting just above $(1,2)$ .
$ff^{-1}(x)=x$ with domain $D_{f^{-1}}=(1,\infty)$ .
The three graphs meet where $f(x)=x$ would be, but here at no common point.

Function, inverse and identityScroll across to see the full diagram.

Answer · (iii)

(2,1)

Explore the idea

f and f⁻¹ with the line y = x

ff⁻¹(x) = x is the part of the dashed line starting just above (1, 1).

Point on f: input xx = 3.00

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))

(3, 2.71828)

Q = (f(x), x)

(2.71828, 3)

on the inverse

Domain of f

x>2

Range of f = domain of f⁻¹

y>1

Midpoint M on y = x

(2.85914, 2.85914)

F for x > 2
F⁻¹ for x > 1
Y = x, which is ff⁻¹ for x > 1
PQ is perpendicular to y = x; its blue midpoint M lies on that line

Key idea

$f$ , $f^{-1}$ and $y=x$ all pass through one point, near $x=3.15$ . That is the fixed point $f(a)=a$ used in part (iv).

Ideas to try

Where do

f

and

f^{-1}

meet the line

y=x

? Compare with part (iv).

(iv)

Equality of composite functions

$a$ is now replaced by a constant for which $ff^{-1}$ and $f^{-1}f$ are equal functions. Explain why $a=f(a)$ , and find $a$ .

Explore this idea

Hint

Both composites equal $x$ , so their domains must match.
$D_{ff^{-1}}=D_{f^{-1}}=R_f$ and $D_{f^{-1}f}=D_f$ .
So $R_f=D_f$ .

Answer · (iv)

a=e^{a-2}

Explore the idea

When does the range start where the domain does?

The domain starts at x = a and the range at f(a). They agree only at a ≈ 3.15, where the graph's start sits on the line y = x.

Start: a = 2

$a=2$ : the range starts at $f(2)=1$ , but the domain starts at 2. They differ, so the composites are not equal.

Marked points at this step, rounded to 6 significant figures.
Point	x	y
start (2, 1)Open point	2	1

F for x > a

Ideas to try

Find the choice of

a

where the start point has equal coordinates.

(v)

Existence of composites

$g(x)=e^{|2-x|}$ for $x\in\mathbb Z$ . Explain why $g^2$ does not exist.

Hint

$g^2$ exists if $R_g\subseteq D_g$ .
Find one output that is not in the domain.

Answer · (v)

g^2

Matching rules are not equal without matching domains

Solution

One-one restrictions

Turning point

Where does the graph stop being one-one?

Inverse functions

Remove the modulus

Make $x$ the subject

Domain

Graphs of inverses and composites

End-points

The composite

f and f⁻¹ with the line y = x

Equality of composite functions

Domains must match

Equate the starting points

Solve

Numerical root

When does the range start where the domain does?

Existence of composites

A counterexample

Turning point

Remove the modulus

Make xxx the subject

Domain

End-points

The composite

Domains must match

Equate the starting points

Solve

Numerical root

A counterexample

Make $x$ the subject