The Purplemath Forums

by **jaybird0827** on Wed Jul 01, 2009 12:02 am

Problem: For what values of $x$ is the curve $y\,=\,e^{-x^2}$ concave downward?

OK, so this calls for a 2nd derivative test. So this is what I did:

$\frac{dy}{dx}\,=\,-2xe^{-x^2}$ . That was easy. But now we have a product.

$\frac{d^2y}{dx^2}\,=\,-2[x(-2xe^{-x^2})\,-\,2e^{-x^2}]\,\,\,\,\,\,\,\,=\,4x^2e^{-x^2}\,+\,4e^{-x^2}$

$\frac{d^2y}{dx^2}\,=\,4e^{-x^2}(x^2\,+1)$ . Now, if I'm looking for concave downward, I want the value of the $y''$ expression to be $<\,0$ ,

$4e^{-x^2}(x^2\,+1)\,<\,0$ . We know $x^2\,+\1\,>\,0$ ; then we must be really looking for $e^{-x^2}\,<\,0$

That's how far I got with it. The answer is supposed to be the interval $[-\frac{\sqrt{2}}{2},\,\frac{\sqrt{2}}{2}]$ .
Did I botch something somewhere?

**Sponsor**

by **Martingale** on Wed Jul 01, 2009 3:58 am

jaybird0827 wrote:Problem: For what values of $x$ is the curve $y\,=\,e^{-x^2}$ concave downward?

OK, so this calls for a 2nd derivative test. So this is what I did:

$\frac{dy}{dx}\,=\,-2xe^{-x^2}$ . That was easy. But now we have a product.

$\frac{d^2y}{dx^2}\,=\,-2[x(-2xe^{-x^2})\,-\,2e^{-x^2}]\,\,\,\,\,\,\,\,=\,4x^2e^{-x^2}\,+\,4e^{-x^2}$

$\frac{d^2y}{dx^2}\,=\,4e^{-x^2}(x^2\,+1)$ . Now, if I'm looking for concave downward, I want the value of the $y''$ expression to be $<\,0$ ,

$4e^{-x^2}(x^2\,+1)\,<\,0$ . We know $x^2\,+\1\,>\,0$ ; then we must be really looking for $e^{-x^2}\,<\,0$

That's how far I got with it. The answer is supposed to be the interval $[-\frac{\sqrt{2}}{2},\,\frac{\sqrt{2}}{2}]$ .
Did I botch something somewhere?

$\frac{d^2y}{dx^2}=-2e^{-x^2}-2x\left(-2xe^{-x^2}\right)=-2e^{-x^2}+4x^2e^{-x^2}$

by **jaybird0827** on Wed Jul 01, 2009 9:08 pm

Martingale wrote:
$\frac{d^2y}{dx^2}=-2e^{-x^2}-2x\left(-2xe^{-x^2}\right)=-2e^{-x^2}+4x^2e^{-x^2}$

I did wonder if I'd messed up on the 2nd derivative. Thanks, Martingale! :clap:

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