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

(a)i

One-one restrictions

f(x)=x2+λx+2f(x)=x^2+\lambda x+2 for real x≤2x\le2, where λ\lambda is a constant. Find the values of λ\lambda for which f−1f^{-1} exists.

Explore this idea

Hint

  1. ff has a minimum at x=−λ2x=-\frac\lambda2.

  2. For x≤2x\le2 to be one-one, the graph must be strictly decreasing on the whole domain: the minimum must be at or beyond x=2x=2.

Answer · (a)i

λ≤−4\lambda\le-4.

Explore the idea

Open interactive exploration

Is the restricted parabola one-one?

Each step shows f for x ≤ 2. The vertex must be at x = 2 or further right for the graph to be strictly decreasing.

Teal: λ = 0. No comparison overlay.

Start: λ = 0

λ=0\lambda=0: vertex at x=0x=0, inside the domain. The graph falls and then rises: not one-one.

  • F on x ≤ 2
  • Vertex or end-point
Ideas to try
  1. Find the smallest λ\lambda shown that still has an inverse.
  2. What does a vertex at x=2x=2 look like? Is it OK?