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

(a)

Rational inequalities

f(x)=(x−7)(x+1)f(x)=(x-7)(x+1) for x>0x>0. Use an algebraic method to solve f(x)<x2−7x2−xf(x)<\dfrac{x^2-7x}{2-x}.

Explore this idea

Hint

  1. Move everything to one side and factorise.

  2. Use x>0x>0 to drop positive factors.

  3. Draw a sign chart for the remaining factors.

Answer · (a)

0<x<20<x<\sqrt2 or 2<x<72<x<7.

Explore the idea

Where is (x − 7)(x² − 2)/(x − 2) negative?

Drag the test value across √2, 2 and 7. The graph is drawn for x > 0.

(x−7)(x−2)(x+2)x−2<0\frac{(x-7)(x-\sqrt2)\left(x+\sqrt2\right)}{x-2}<0

Choose an interval to test it. Shading marks the solution intervals.

Intervalx − √2x − 7x − 2product
negativenegativenegativenegative
positivenegativenegativepositive
positivenegativepositivenegative
positivepositivepositivepositive
Test value of xx = 1.00
Try an example

(x − √2)−(x − 7)−(x − 2)−⇒ product negative

Is it < 0?
Yes
x is in the solution set

The shaded strips are where the expression satisfies the inequality; they match the green-tinted columns of the chart.

  • Expression, and where it is negative
  • Your test value
Key idea

The factor x+2x+\sqrt2 is always positive for x>0x>0: drop it from the chart, not from the working.

Ideas to try
Test x=1x=1, x=1.8x=1.8, x=3x=3 and x=8x=8.
(b)i

One-one functions

Determine whether f−1f^{-1} exists.

Explore this idea

Hint

  1. Find two inputs with the same output.

Answer · (b)i

No inverse: f(1)=f(5)=−12f(1)=f(5)=-12.

Explore the idea

The horizontal line test

Below the open point (0, −7) the line meets the graph twice.

Height of the horizontal linek = -12
Try an example
Test boundary heights

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

Number of distinct intersections
2

Two intersections, both with 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
1-12
5-12
  • F
  • Y = k
  • Intersections
  • Intersection selected in the coordinate table
Key idea

One counterexample suffices: f(1)=f(5)f(1)=f(5).

Ideas to try
Why is there only one intersection when k>−7k>-7?
(b)ii

Range of a composite

gg is one-one with domain (0,∞)(0,\infty) and range (4,∞)(4,\infty). Find the range of fgfg.

Explore this idea

Hint

  1. Rg=(4,∞)R_g=(4,\infty) is the restricted domain for ff.

  2. Read the range of ff on (4,∞)(4,\infty) from the graph.

Answer · (b)ii

Rfg=(−15,∞)R_{fg}=(-15,\infty).

Explore the idea

Only the right-hand arm of f is used

g feeds f only numbers above 4, so only the increasing part of the parabola counts.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • F on (4, ∞)
Ideas to try
What would the range be if RgR_g were (2,∞)(2,\infty)?
(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.

(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

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.