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

(i)

Range

For all real xx, f(x)=21+(5x−1)2f(x)=\dfrac{2}{1+(5x-1)^2}. Find its range.

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Hint

  1. The denominator is smallest when 5x−1=05x-1=0.

  2. As ∣x∣→∞|x|\to\infty the fraction tends to 0 but never reaches it.

Answer · (i)

Rf=(0,2]R_f=(0,2].

Explore the idea

Open interactive exploration

A bell curve with maximum 2

f is symmetric about x = 1/5.

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
  • Axis of symmetry
Ideas to try
Read off the range: which yy-values does the curve take?