Matching rules are not equal without matching domains

The function ff is defined by f:x↦e∣2−x∣ f:x\mapsto e^{|2-x|}, x∈Rx\in\mathbb R, x>ax>a, where aa is a constant. It is given that aa is the least possible value such that f−1 f^{-1} exists.

(i) State the value of aa. mark

(ii) Hence find f−1(x) f^{-1}(x) and state the domain of f−1 f^{-1}. marks

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

It is now given that aa is replaced by a constant such that the functions ff−1 ff^{-1} and f−1ff^{-1}f are equal.

(iv) Explain why the new value of aa satisfies the equation a=f(a) a=f(a), and find aa. marks

(v) The function gg is defined by g:x↦e∣2−x∣ g:x\mapsto e^{|2-x|}, x∈Zx\in\mathbb Z. Explain why composite g2 g^2 does not exist. mark

[2021/EJC/Promo/8]

Total marks: 11

Solution

If this part uses an earlier result, revisit (ii) Inverse functions or open all working together.

(iii)

Graphs of inverses and composites

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

Explore this idea

Hint

  1. ff: starts just right of (2,1)(2,1) and rises. f−1f^{-1}: reflection, starting just above (1,2)(1,2).

  2. ff−1(x)=xff^{-1}(x)=x with domain Df−1=(1,∞)D_{f^{-1}}=(1,\infty).

  3. The three graphs meet where f(x)=xf(x)=x would be, but here at no common point.

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

Answer · (iii)

Open endpoints (2,1)(2,1) for f, (1,2)(1,2) for inverse, (1,1)(1,1) for ff−1(x)=xff^{-1}(x)=x.

Explore the idea

Open interactive exploration

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
Reset the controls

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>2x>2
Range of f = domain of f⁻¹
y>1y>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

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

Ideas to try
Where do ff and f−1f^{-1} meet the line y=xy=x? Compare with part (iv).