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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

multipledifferentones
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1 Answers1

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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.

Sarvesh Ravichandran Iyer
  • 76,105