# Intermediate Value Theorem Confusion

I have a question on this online website I'm trying to learn calculus on. What I am really confused about the Intermediate Value Theorem is: it says I should set it to 0 but I'm totally lost at the steps used to approach and take this? Perhaps someone can guide me in the right direction.

Do I plug in 1 and 0 for $x$ and then compare it?

Use the Intermediate Value Theorem to show that there is a solution to the equation $e^{-x^2} = x$ in the interval (0, 1).

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Are you sure that your statement is correct? $e-x^2=x$ has no solution in $(0,1)$. – JBC Jun 17 '12 at 21:10
Did you mean, perhaps $e^{-x^2} = x$? – Arturo Magidin Jun 17 '12 at 21:12
Oops you are right – soniccool Jun 17 '12 at 21:19

The Intermediate Value Theorem states:

Theorem. Let $f(x)$ be a function that is continuous on $[a,b]$. If $k$ is a number between $f(a)$ and $f(b)$, then there exists (at least) one $c$, $a\lt c \lt b$, such that $f(c) = k$.

I often tell my Calculus students to think of the Intermediate Value Theorem as the "Goldilocks Theorem". You are trying to find a point where $f$ takes the value $k$. If one of $f(a)$ and $f(b)$ is "too much", and the other is "too little", then somewhere in between is a point where $f$ is "just right."

The most common (but by no means only) use of the Intermediate Value Theorem is to find points where $f(x)$ is equal to $0$. If you can find a point where $f(x)$ is positive and one where $f(x)$ is negative, then somewhere in between there will be a point where $f(x)$ is exactly $0$ (provided $f$ is continuous).

In the case at hand, you want to find a solution to $e^{-x^2}=x$. So we can take $f(x) = e^{-x^2}-x$; we are looking for a point where $f(x)=0$. This function is continuous everywhere. Since $f(0) = e^{0}-0 = 1$ and $f(1) = e^{-1}-1 \lt 0$, we have:

$f(x)$ is continuous on $[0,1]$, it is positive at $0$ (value is "too much"), and it is negative at $1$ (value is "too little"). Therefore, by the Intermediate Value Theorem, there is at least one point $c$ somewhere in $(0,1)$ where $f(c)=0$ ("just right").

But if $f(c)=0$, then that means that $0 = e^{-c^2}-c$, so $e^{-c^2}=c$. That is, the equation $e^{-x^2} = x$ has at least one solution in $(0,1)$.

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I assume that you want to resolve $e^{-x^2}=x$ (see the comments).
So let $f(x)=e^{-x^2}-x$, $f$ is continuous, $f(0)=1$, $f(1)=e^{-1}-1<0$ so by the IVT, it exists $x_0\in(0,1)$ such as $f(x_0)=0$, ie $e^{-x_0^2}=x_0$.

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