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 (iii) Graphs of inverses and composites or open all working together.

(iv)

Equality of composite functions

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

Explore this idea

Hint

  1. Both composites equal xx, so their domains must match.

  2. Dff−1=Df−1=RfD_{ff^{-1}}=D_{f^{-1}}=R_f and Df−1f=DfD_{f^{-1}f}=D_f.

  3. So Rf=DfR_f=D_f.

Answer · (iv)

a=ea−2a=e^{a-2}; admissible root a=3.15a=3.15 (3 s.f.).

a≥2a\ge2; more precisely 3.14619….

Explore the idea

Open interactive exploration

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.

Teal: a = 2. No comparison overlay.

Start: a = 2

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

  • F for x > a
Ideas to try
Find the choice of aa where the start point has equal coordinates.