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

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?
(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

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)?
(b)i

Existence of composites

g(x)=x2+x+1g(x)=x^2+x+1 for all real xx and h(x)=x+4xh(x)=x+\frac4x for real x>0x>0. Show that hghg exists.

Explore this idea

Hint

  1. Find the range of gg by completing the square.

  2. Compare with Dh=(0,∞)D_h=(0,\infty).

Answer · (b)i

hghg exists: Rg=[3/4,∞)⊆Dh=(0,∞)R_g=[3/4,\infty)\subseteq D_h=(0,\infty).

Explore the idea

Do the outputs of g stay positive?

Drag x (shown over [−3, 2]; g has domain ℝ). The range of g starts at ¾, well inside D_h = (0, ∞).

Input x (in the domain of g)x = -0.50
Try an example
input
-0.5
↓ g
g(x)
0.75
↓ h
hg(x)
6.08333

0.75 is an allowed input of h.

D_hR_g048

D_h: (0, ∞); R_g: [0.75, ∞). Finite bounds are shown to 6 significant figures. Hollow ends are excluded; arrows continue beyond this window.

Selected output: 0.75.

The composite exists

Rg=[34,∞)⊆Dh=(0,∞)R_g=[\tfrac34,\infty)\subseteq D_h=(0,\infty): even the smallest output 34\frac34 is positive.

  • G(x)
  • Hg(x)
  • Lower end of D_h (open at 0)
Key idea

gg's outputs never reach 0, so hh is never asked to divide by zero.

Ideas to try
Is there any xx for which g(x)≤0g(x)\le0?
(b)ii

Range of a composite

Find the range of hghg.

Explore this idea

Hint

  1. The range of gg becomes the restricted domain of hh.

  2. Sketch h(x)=x+4xh(x)=x+\frac4x on [34,∞)\left[\frac34,\infty\right) and read the range.

Answer · (b)ii

Rhg=[4,∞)R_{hg}=[4,\infty).

Explore the idea

Feed a whole interval into h

The range of g becomes the input domain of h. Move its lower bound to see when the interior minimum is lost.

In this question g(x) = x² + x + 1 reaches [3/4, ∞), so hg uses h on that interval. The original-composite preset restores this exact range; other boundaries are what-if experiments.

Lower bound supplied to h0.75

Closed endpoint: the boundary is included unless the original domain excludes it.

Try an example

Proposed input domain

[0.75, ∞)

Proposed input domain036

Retained inputs

[0.75, ∞)

Retained inputs036

Reached outputs

[4, ∞)

Reached outputs0510

Original allowed inputs

(0, ∞)

These exclusions stay in force when a boundary changes. Arrows indicate continuation beyond the drawing window.

The input interval contains t = 2, where h reaches 4. Its left endpoint does not determine the minimum.

The graph shows a finite window; arrows and interval notation describe the full set. Decimal endpoints are rounded.