Restrict the domain to choose an inverse

For all real x, f(x)=2/[1+(5x−1)2]f(x)=2/[1+(5x-1)^2].

(i) Find its range. marks

(ii) Explain why it has no inverse. mark

(iii) Restrict its domain to x≥kx\ge k: find the least k that gives an inverse, and find that inverse. marks

(iv) Sketch the restricted f and its inverse together, clearly showing their relationship. marks

(v) Show algebraically that a solution of f(x)=f−1(x)f(x)=f^{-1}(x) satisfies 25x3−10x2+2x−2=025x^3-10x^2+2x-2=0. marks

[2011/IJC/Prelim/I/9]

Total marks: 10

Solution

If this part uses an earlier result, revisit (iii) Restricting a domain or open all working together.

(iv)

Graph of an inverse

Sketch the restricted ff and f−1f^{-1} on the same diagram, clearly showing their relationship.

Explore this idea

Hint

  1. Use the same scale on both axes.

  2. Draw ff from (15,2)\left(\frac15,2\right) downwards, and reflect it in y=xy=x.

  3. Mark the end-points.

Restricted f and f⁻¹: equal coordinate units
Restricted f and f⁻¹: equal coordinate unitsScroll across to see the full diagram.

Answer · (iv)

Reflection in y=xy=x; endpoints (1/5,2),(2,1/5)(1/5,2),(2,1/5).

Equal coordinate-unit scales.

Explore the idea

Open interactive exploration

Restricted f and its inverse

Drag the input. P = (x, f(x)) is on f and Q = (f(x), x) is on f⁻¹.

Point on f: input xx = 0.50
Reset the controls

Reflection swaps the coordinates. The segment PQ is perpendicular to y = x, and its blue midpoint M lies on that line.

P = (x, f(x))
(0.5, 0.61538)
Q = (f(x), x)
(0.61538, 0.5)
on the inverse
Domain of f
x≥15x\ge\frac15
Range of f = domain of f⁻¹
0<y≤20<y\le2
Midpoint M on y = x
(0.55769, 0.55769)
  • F restricted to x ≥ 1/5
  • F⁻¹
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

The inverse's vertical asymptote x=0x=0 is the mirror of ff's horizontal asymptote y=0y=0.

Ideas to try
Where do ff and f−1f^{-1} meet? Is it on y=xy=x?