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)iii Proving one-one or open all working together.

(b)iv

Inverse values

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

Explore this idea

Hint

  1. (fg)−1(20)=x(fg)^{-1}(20)=x means fg(x)=20fg(x)=20.

  2. Expand f(g(x))f(g(x)) as a quadratic in g(x)g(x).

  3. Reject the root that lies outside RgR_g.

Answer · (b)iv

(fg)−1(20)=ln⁡(9/4)(fg)^{-1}(20)=\ln(9/4).

Explore the idea

Open interactive exploration

Which x gives fg(x) = 20?

fg(x) = (4eˣ − 7)(4eˣ + 1) is increasing for x > 0, so every height above −15 is hit exactly once.

Height of the horizontal linek = 20
Try an example
Test boundary heights

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

Number of distinct intersections
1

Exactly one intersection, so the composite is one-one on x > 0.

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.
Pointxy
0.8109320
  • Fg
  • Y = k
  • Input x
  • Intersection selected in the coordinate table
Key idea

A single intersection per height is exactly what 'one-one' means graphically.

Ideas to try
Set k=20k=20: the intersection should be at x≈0.81=ln⁡94x\approx0.81=\ln\frac94.