I am a HS student and currently learning Rolle's Theorem.
I have gotten the question:
Prove that there are exactly two positive real numbers $x$ such that $e^x = 3x$.
This is what I have done to answer:
$f(0) = 1 > 0$,
$f(1) = e – 3 < 0$.
There must be a point between $x=1$ and $x=0$, $x_{0}$, such that $f(x_{0}) = 0$. Therefore there is one root.
Suppose that there is another root such that $x_{1} > x_{0}$.
By Rolle's Theorem, as this function is differentiable and continuous, there must be a point $c$, such that $f'(c) = 0$ between $x = x_{1}$ and $x = x_{0}$.
$f'(x) = e^x – 3$.
This can equal $0$, but there is only one root to this equation. Therefore, there can only be one other root as there is one turning point.
I am not sure if this is sufficient or actually legitimate…
Does this work or should more be added/made more clear? It is just a question from my textbook so does not need to be perfect but needs to show the point and be pretty correct.
Many thanks,
Aidanaidan12
Best Answer
Actually, you require two real roots of this expression. You have only one with you. The Rolle argument which you have done , proves that there can be at most two real roots.
You have not found the other real root. To do that, note that $f(2) = e^2 - 6 > 0$. So there is a real root between $1$ and $2$ as well. These can be the only real roots by Rolle. So you were almost done, but needed the location of the second root.
Remember : By Rolle's theorem, if the derivative has at most $n$ real roots, then the function itself has at most $n+1$ real roots. You cannot say anything more : so for example, seeing that the derivative is zero does not tell you that there needs to be a second root.