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)ii Inverse with a restricted domain or open all working together.

(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

Open interactive exploration

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?