mycroft holmes
Last Activity: 14 Years ago
If x1 /= x2 , then by Lagrange's Mean Value Theorem, we have f(x2) - f(x1)/(x2-x1) = f'(c) for some c in the interval (x1 , x2).
Since f'(c)>0, and x2 > x1 , this implies f(x2) > f(x1). Thus, we have that if x1 /= x2 , then f(x2) /= f(x1).
Thus f'(x)>0 proves that f(x) is one-one
In general if f(x) is monotonic, it is also one-one.