Map the whole interval, not only its endpoints

(a) f(x)=x2+λx+2f(x)=x^2+\lambda x+2 for real x≤2x\le2, with constant lambda.

(i) Find the values of lambda for which an inverse exists. marks

(ii) With λ=−4\lambda=-4, define the inverse, including its domain. marks

(b) g(x)=x2+x+1g(x)=x^2+x+1 for all real x, and h(x)=x+4/xh(x)=x+4/x for real x>0x>0.

(i) Show hg exists. marks

(ii) Find its range. marks

[modified 2011/HCI/Prelim/II/2]

Total marks: 9

Solution

If this part uses an earlier result, revisit (a)i One-one restrictions or open all working together.

(a)ii

Inverse with a restricted domain

With λ=−4\lambda=-4, define f−1f^{-1}, including its domain.

Explore this idea

Hint

  1. Complete the square: y=(x−2)2−2y=(x-2)^2-2.

  2. Solve for xx and choose the branch with x≤2x\le2.

  3. The domain of f−1f^{-1} is RfR_f.

Answer · (a)ii

f−1(x)=2−x+2f^{-1}(x)=2-\sqrt{x+2}, x≥−2x\ge-2.

Explore the idea

Open interactive exploration

The left half of a parabola and its inverse

f(x) = x² − 4x + 2 for x ≤ 2. Q = (f(x), x) lies on the inverse 2 − √(x + 2).

Point on f: input xx = 0.00
Try an example

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, 2)
Q = (f(x), x)
(2, 0)
on the inverse
Domain of f
x≤2x\le2
Range of f = domain of f⁻¹
y≥−2y\ge-2
Midpoint M on y = x
(1, 1)
  • F restricted to x ≤ 2
  • F⁻¹
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

The inverse starts at (−2,2)(-2,2), the mirror of the vertex (2,−2)(2,-2). Its domain starts at −2-2.

Ideas to try
Move xx to 2. Where is QQ? What is f−1(−2)f^{-1}(-2)?