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 $x \in ]3,4[$ ?

  • $\begingroup$ Do you mean using a combination of Bolzano's theorem and Rolle's theorem? $\endgroup$ – mfl Jun 16 '14 at 13:33
  • $\begingroup$ Only with Bolzano's theorem. I found the solution in the first answer: i need to prove that f is strictly monotonically decreasing. $\endgroup$ – Jorge Jun 16 '14 at 13:37

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.