# Using Rolle's Theorem to prove that there are two roots to a function.

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

• you could simply also note that $f(2) > 0$, so... – ab123 Dec 8 '19 at 11:43

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.