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Considering the Bolzano's theorem: Let $f$ be continuous at each point of a closed interval $[a, b]$ and assume that $f(a)$ and $f(b)$ have opposite signs. Then there is at least one $c$ in the open interval $(a, b)$ such that $f(c) = 0$.

How can I use the Bolzano's theorem to prove that exists only one solution?

For example:

$f(x) = 25 + 60 e^{-0.35x}$

how can I prove that exists only one solution for $f(x)=40$ in the interval $x \in ]3,4[$ ?

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Do you mean using a combination of Bolzano's theorem and Rolle's theorem? –  mfl Jun 16 at 13:33
    
Only with Bolzano's theorem. I found the solution in the first answer: i need to prove that f is strictly monotonically decreasing. –  Jorge Jun 16 at 13:37

2 Answers 2

up vote 11 down vote accepted

If the function $f$ is in addition strictly monotonically increasing (or decreasing as your example), then there is exactly one solution.

The Bolzano theorem alone cannot help to prove uniqueness.

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Check the signs of f(x)-40 at the endpoints of that interval to conclude they are different from each other. Then notice that f(x) is strictly decreasing (f'(x)<0) as the part e^-c*x is strictly decreasing and the constant doesn't affect this. Combine these two bits of information to conclude that only one solution exists.

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