A quarter ellipse becomes a self-inverse circle

Sketch the curve y2+ax2=5 y^2+ax^2=5 for x≥0 x\ge0, y≥0 y\ge0, where aa is positive. mark

f:x↦5−ax2 f:x\mapsto\sqrt{5-ax^2}, x∈Rx\in\mathbb R, 0≤x≤5/a0\le x\le\sqrt{5/a}; g:x↦1+e−x g:x\mapsto1+e^{-x}, x∈Rx\in\mathbb R, x≥0 x\ge0.

Show that f−1 f^{-1} exists and define f−1 f^{-1} in a similar form. marks

Given f2(x)=xf^2(x)=x for all x∈Rx\in\mathbb R, 0≤x≤5/a0\le x\le\sqrt{5/a},

(i) show a=1 a=1 without evaluating f2(x) f^2(x), marks

(ii) using (i), show fgfg exists and find its corresponding range. marks

[2011/ACJC/Prelim/I/9]

Total marks: 10

Solution

If this part uses an earlier result, revisit (i) Self-inverse by symmetry or open all working together.

(ii)

Range of a composite

g(x)=1+e−xg(x)=1+e^{-x} for x≥0x\ge0. Show that fgfg exists and find its range.

Explore this idea

Hint

  1. With a=1a=1, Df=[0,5]D_f=[0,\sqrt5].

  2. Find RgR_g and compare.

  3. Feed RgR_g through ff.

Answer · (ii)

fgfg exists and Rfg=[1,2)R_{fg}=[1,2).

Explore the idea

Open interactive exploration

g's outputs lie inside [0, √5]

Drag x. The range (1, 2] of g sits entirely within the domain [0, √5] of f.

Input x (in the domain of g)x = 1.00
Try an example
input
1
↓ g
g(x)
1.36788
↓ f
fg(x)
1.76887

1.36788 is an allowed input of f.

D_fR_g01.53

D_f: [0, 2.23607]; R_g: (1, 2]. Finite bounds are shown to 6 significant figures. Hollow ends are excluded; arrows continue beyond this window.

Selected output: 1.36788.

The composite exists

Rg=(1,2]⊆Df=[0,5]R_g=(1,2]\subseteq D_f=[0,\sqrt5], since 2<52<\sqrt5.

  • G(x)
  • Fg(x)
  • Top of D_f
Key idea

fgfg is increasing: gg falls and ff falls, so the composite rises from f(2)=1f(2)=1 towards f(1)=2f(1)=2.

Ideas to try
As xx grows, what happens to g(x)g(x) and fg(x)fg(x)?