There is an indirect proof for a function having only one root in a given interval, which involves the use of the Mean Value Theorem. Allow me to illustrate this with an example:
Let: f(x) = x^2 + 2x.cos(x) - 1
a) Show that f'(x) > 0
f'(x) = 2x + 2cos(x) - 2xsin(x) = 2x(1-sin(x)) + 2cos(x)
0 <= sin(x) <= 1
-1 <= -sin(x) <= 0
0 <= 1-sin(x) <= 1
cos(x) > 0
b) Using the Intermediate Value Theorem show that f(x) has a root in the range (0,1).
f(0) = -1
f(1) = 2cos1 > 0
f(x) is continuous for this interval and it's value goes from -ve to +ve: Thus by the Intermediate Value Theorem it must have at least one root in the said interval.
c) Using the Mean Value Theorem show that f(x)=0 has only one root in the interval (0,1)
We shall prove this by contradiction. (See: Proof By Contradiction).
Assume that f(x) has 2 roots, c1 and c2 in the given interval.
f(c1) = f(c2) = 0
By the mean value theorem we have:
f'(c).(c2 - c1) = f(c2) - f(c1)
f'(c) = 0
But in part a it was shown that f'(c) > 0 in the range (a,b)
We have a contradiction, and thus the original assumption that there are two roots, has been disproven.
See also: Intermediate Value Theorem, Mean Value Theorem