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

Open interactive exploration

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.