Expert: Ahmed Salami - 10/31/2009

Question
f(x)= a-x/(a+x) where a is some real number that we can choose. for what value of a is f(x) its own inverse.

Answer
Hi Vsikandari,
f(x) = a-x/a+x
Let f(x) be y and its inverse z
y = a-x/a+x
y(a+x) = a-x
ay + xy = a - x
xy + x = a - ay
x(y+1) = a(1-y)
x = a(1-y)/(y+1)
And so, the inverse of f(x)
z = a(1-x)/(x+1)
When f(x) = z
(a-x)/(a+x) = a(1-x)/(x+1)
(a-x)(x+1) = a(1-x)(a+x)
ax + a - x² - x = a(a + x - ax - x²)
ax + a - x² - x = a² + ax - a²x - ax²
a - x² - x = a² - a²x - ax²
a - x² - x - a² + a²x + ax² = 0
(a-1)x² + (a²-1)x + a(1-a) = 0
The above would only always be true for a value of a that makes all of (a-1), (a²-1) and a(a-1) equal to 0 at the same time.
For a-1 = 0
a = 1
For a²-1 = 0
a = 1 or -1
For a(1-a) = 0
a = 0 or 1
The value we need is therefore a = 1

We could also have said, from (a-1)x² + (a²-1)x + a(1-a) = 0
(a-1)[x² + (a+1)x - a] = 0
i.e a - 1 = 0
a = 1

Regards