Calculus option question

[haese225]

Calc. option question:

Q: f(x) = e^x - 3x = 0, prove that there are exactly two positive real numbers x.

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My working:

Since, f(x) is continuous and differentiable, I thought about how I could apply Rolle's theorem 

f'(x) = e^x - 3 = 0

=> x = ln 3 

Here, f(ln 3) = 3 - 3ln 3 < 0 

I reasoned: Since, there is only one stationary point (x = ln 3) and this occurs when f(x) < 0, therefore there must be exactly two roots. 

But how do we show that they are both positive? 

Thanks all! 

[SC2Player]

The converse of Rolle's theorem is not necessarily true, so you can't really use it as you outlined here.  A counterexample is f(x)=-x^2-1 - it has exactly one stationary point too, below zero, but it has no real roots.  Another counterexample is f(x)=x^3 - 1 - has one stationary point, but only one real root.  

What you can apply is the intermediate value theorem (might be worthwhile to show that f(x) is continuous everywhere).  You've got the stationary point already, which is at x=ln 3, and shown that this is the only stationary point.  You've also shown that this is less than 0.  

Now let's take the points (0, f(0)) and (0, f(2)).  0<ln 3, so (0, f(0)) is to the left of (ln 3, f(ln 3)).  f(0)=e^0-3*0=1>0.  As f(0)>0 and f(ln3)<0, there must be at least one point between (0, f(0)) and (ln 3, f(ln 3)) where f(x)=0.  However, for there to be more than one point where f(x)=0, there must be another stationary point (can you see why?).  You've already shown that there is only one stationary point in the function, so there must be exactly one root here only.  As this point is to the right of (0, f(0)), the root must be positive.  

Apply the same idea to f(2), and you've shown that there are two positive roots to the function f(x).  Prove that the function is strictly increasing for x>ln 3, and strictly decreasing for x<ln 3, and you have proven that these are the only two positive roots. 

I think this is kind of what you were trying to do, but it's just a bit of an issue with the theorem that you're applying.  

Edited July 12, 2017 by SC2Player