A non-invertible function can have an invertible composite

f:x↦(x−7)(x+1) f:x\mapsto(x-7)(x+1), x>0 x>0.

(a) Use an algebraic method to find exact solution of f(x)<(x2−7x)/(2−x) f(x)<(x^2-7x)/(2-x). marks

(b)(i) Determine if f−1 f^{-1} exists. marks

Given gg is one-one with domain (0,∞)(0, \infty) and range (4,∞)(4, \infty):

(ii) Find range of fgfg. marks

(iii) A function hh is one-one if for any α,β∈Dh\alpha, \beta\in D_h , h(α)=h(β) h(\alpha)=h(\beta) implies α=β\alpha=\beta . Show that fgfg is one-one. marks

(iv) If g(x)=4exg(x)=4e^x for x>0 x>0, find exact (fg)−1(20)(fg)^{-1}(20). marks

[2011/RI/Promo/13]

Total marks: 15

Solution

If this part uses an earlier result, revisit (b)ii Range of a composite or open all working together.

(b)iii

Proving one-one

A function hh is one-one if h(α)=h(β)h(\alpha)=h(\beta) implies α=β\alpha=\beta. Show that fgfg is one-one.

Hint

  1. Assume fg(α)=fg(β)fg(\alpha)=fg(\beta).

  2. Both g(α)g(\alpha) and g(β)g(\beta) lie in (4,∞)(4,\infty), where ff is one-one.

  3. Then use that gg is one-one.

Answer · (b)iii

fgfg is one-one: equal outputs force g(α)=g(β)g(\alpha)=g(\beta) and hence α=β\alpha=\beta.

The alternative factor g(α)+g(β)−6 is positive because both g-values exceed 4.