2

Problem: Prove using the Intermediate Value Theorem that $ f(x) = e^x $ and $ g(x) = x^e $ intersect.

My attempt:

First, here is the definition of the IVT that I am familiar with - "Suppose $f$ is continuous over $[a,b]$ and $f(a) \neq f(b)$. If $Y$ is strictly between $a$ and $b$, then there exists a $c \in (a,b)$ such that $f(c)=Y$."

However, in this problem, $f(x)=e^x \geq g(x)=x^e$ for all $x$.

So if we let $h(x)=e^x - x^e$, we get a function that is never negative, but touches the x-axis once where $f(x)=g(x)$.

Since, in the IVT, $c$ cannot be one of the endpoints, I have tried to think of other ways to approach the problem.

I wondered if it would be okay to prove that $h'(x)=0$ where the functions touch, because $h'$ actually does cross the x-axis, but it crosses at two points, $x=1$ and $x=e$, so I don't know if it would work.

I also thought about using a piecewise function.

$$ j(x) = \begin{cases} e^x-x^e, x \leq e\newline x^e-e^x, x > e \end{cases} $$

And now $j(x)$ crosses the x-axis and I could go about proving that using the IVT. But again, I'm not sure if this is breaking some hidden rule or something.

Are either of my approaches okay? Or is there another direction I should be going? Thanks.

kai
  • 75
  • 4
    The intermediate value theorem does not apply here as you correctly pointed out , but we can easily guess the root $x=e$. The exercise is ill-posed. The functions do not intersect , they touch each other. – Peter Aug 25 '23 at 21:36
  • There’s nothing more difficult than a proof of the obvious. We see right away that they are tangent at x=e theres no need for the IVT. If you asked to prove that they intersect using the fundamental theorem of arithmetic I think the problem wouldn’t be any easier. – Lee Fisher Aug 25 '23 at 21:47

1 Answers1

1

You remark that function $e^x$ is always greater than $x^e$. We know function $e^x$ is always crescent. Since we intuit less on irrationals, function $x^e$ is less intuitive: its derivative is $e \cdot x^{e-1}$, which is always positive for $x \in \mathbb{R}$ (also not very intuitive, right!?). Therefore, it is something between functions $x^2$ and $x^3$, closer to the latter. Take a look at: https://www.wolframalpha.com/input?i=plot+real+part+x%5Ee

So far, so good. Since we support on above visualization, the intersection occurs for $x>0$ and if it exists, occurs at least once, because both functions are strictly positive on this interval.

Let us use check control points 0, 1, 2, 3, 4:

$$(f(0), g(0))=(1, 0)$$ $$(f(1), g(1))=(e, 1)$$ $$(f(2), g(2))=(e^2, 2^e) \approx (9, 8)$$ $$(f(3), g(3))=(e^3, 3^e) \approx (27, 27)$$ $$(f(4), g(4))=(e^4, 4^e) \approx (81, 64)$$

Wait, what!? Somewhere between 2 and 3, they got really close and then $e^x$ started growing again. Therefore, our best chance finding the intersection point(s) is between 2 and 3.

Bruno Lobo
  • 496