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 Inverse Inverse functions or open all working together.

(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

Open interactive exploration

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?