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, c_{1} and c_{2} in the given interval.

f(c_{1}) = f(c_{2}) = 0

By the mean value theorem we have:

f'(c).(c_{2} - c_{1}) = f(c_{2}) - f(c_{1})

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