Let \(f(x) = \frac{ax}{x + 1}.\) Find the constant \(a\) so that \((f(x)) = x \) for all \(x \neq -1\)
I found that a=x+1 (im not sure if that's right but the question wants a constant for a. Help!
I really would appreciate it guys ;)
f(x) = x
ax = (x + 1) x
ax = x^2 + x
x^2 + ( 1 - a)x = 0
x ( x + (1 - a) ) = 0 ⇒ x = 0 (reject) or
x + (1 - a) = 0
x = - (1 - a)
x = a - 1
x + 1 = a
This must be correct because
ax (x + 1)x
____ = _______ = x
x + 1 x + 1
