# Math Help - Intermediate Value Theorem

1. ## Intermediate Value Theorem

Dont really understand how to do this problem.

Use the Intermediate Value Theorem to show that there is a root of the equation in the given interval.

e^(-x^2)) = x (0,1)

2. Let $f(x)=e^{-x^2}-x$, observe that $f$ is a continuous function. Since $f(0)=1$ and $f(1)<0$,(because $e>1\Rightarrow e^{-1}<1 \Rightarrow e^{-1}-1<0$) thanks to IVT, $f(x)=0$ in $(0,1)\Rightarrow e^{-x^2}=x$ there.

3. Originally Posted by TastyBeverage
Dont really understand how to do this problem.

Use the Intermediate Value Theorem to show that there is a root of the equation in the given interval.

e^(-x^2)) = x (0,1)
Hint: Let $f(x) = e^{-x^2} - x$. Now $f(0) > 0$ and $f(1) < 0$.

EDIT: Somebody been faster.