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

Sketch

Sketching conics

Sketch y2+ax2=5y^2+ax^2=5 for x≥0x\ge0, y≥0y\ge0, where a>0a>0.

Explore this idea

Hint

  1. Divide by 5: y25+x25/a=1\frac{y^2}5+\frac{x^2}{5/a}=1 is an ellipse.

  2. Find the intercepts and keep the first quadrant.

First quadrant ellipse
First quadrant ellipseScroll across to see the full diagram.

Answer · Sketch

Quarter-ellipse endpoints (0,5),(5/a,0)(0,\sqrt5),(\sqrt{5/a},0).

a>0a>0, x,y≥0.

Explore the idea

The quarter-ellipse changes with a

The vertical intercept stays at √5. The horizontal intercept √(5/a) shrinks as a grows.

Teal: a = 0.5. No comparison overlay.

Start: a = 0.5

a=0.5a=0.5: intercepts (0,5)(0,\sqrt5) and (10,0)≈(3.16,0)(\sqrt{10},0)\approx(3.16,0).

  • Quarter-ellipse
  • Intercepts
Ideas to try
Which value of aa makes the curve look the same after swapping xx and yy?
Inverse

Inverse functions

f(x)=5−ax2f(x)=\sqrt{5-ax^2} for 0≤x≤5/a0\le x\le\sqrt{5/a}. Show that f−1f^{-1} exists and define it in a similar form.

Hint

  1. Show ff is one-one: any horizontal line meets the graph at most once.

  2. Rearrange y=5−ax2y=\sqrt{5-ax^2} for xx.

  3. Df−1=RfD_{f^{-1}}=R_f.

Answer · Inverse

f−1(x)=(5−x2)/af^{-1}(x)=\sqrt{(5-x^2)/a}, 0≤x≤50\le x\le\sqrt5.

Strictly decreasing original f is one-one.

(i)

Self-inverse by symmetry

Given f2(x)=xf^2(x)=x for all xx in the domain, show that a=1a=1 without evaluating f2(x)f^2(x).

Explore this idea

Hint

  1. f2(x)=xf^2(x)=x means f=f−1f=f^{-1}.

  2. A function equal to its inverse has equal domain and range, Df=RfD_f=R_f.

Answer · (i)

a=1a=1.

Identity on the entire domain forces Df=Rf.

Explore the idea

A quarter circle is its own inverse

With a = 1 the graph is √(5 − x²): symmetric about y = x.

Point on f: input xx = 1.00
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))
(1, 2)
Q = (f(x), x)
(2, 1)
on the inverse
Domain of f
0≤x≤50\le x\le\sqrt5
Range of f = domain of f⁻¹
0≤y≤50\le y\le\sqrt5
Midpoint M on y = x
(1.5, 1.5)
  • F with a = 1
  • F⁻¹, the same curve
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

Equal domain and range ([0,5][0,\sqrt5] both) is what forces a=1a=1.

Ideas to try
Move xx: are PP and QQ on the same curve?
(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

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)?