Intermediate Value Theorem

**TastyBeverage** · February 8th 2009, 07:41 PM

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)

**Abu-Khalil** · February 8th 2009, 08:19 PM

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.

**ThePerfectHacker** · February 8th 2009, 08:19 PM

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.

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