Page 1 of 1

### For what values of x is the curve y=e^(-x^2) ...?

Posted: 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?

### Re: For what values of x is the curve y=e^(-x^2) ...?

Posted: Wed Jul 01, 2009 3:58 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?
$\frac{d^2y}{dx^2}=-2e^{-x^2}-2x\left(-2xe^{-x^2}\right)=-2e^{-x^2}+4x^2e^{-x^2}$

### Re: For what values of x is the curve y=e^(-x^2) ...?

Posted: Wed Jul 01, 2009 9:08 pm

$\frac{d^2y}{dx^2}=-2e^{-x^2}-2x\left(-2xe^{-x^2}\right)=-2e^{-x^2}+4x^2e^{-x^2}$
I did wonder if I'd messed up on the 2nd derivative. Thanks, Martingale!