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 (b)i Existence of composites or open all working together.

(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

Open interactive exploration

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.