Question

determine the interval(s) on which the function (y(t)=e^{-\frac{t^{2}}{12}}) is concave down. concave down:

Explanation:

Step1: Find the first - derivative

Let $y(t)=e^{-\frac{t^{2}}{2}}$. Using the chain - rule, if $u =-\frac{t^{2}}{2}$, then $y = e^{u}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=e^{u}$, and the derivative of $u$ with respect to $t$ is $\frac{du}{dt}=-t$. So, by the chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}=-te^{-\frac{t^{2}}{2}}$.

Step2: Find the second - derivative

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u=-t$ and $v = e^{-\frac{t^{2}}{2}}$. The derivative of $u$ with respect to $t$ is $u^\prime=-1$, and the derivative of $v$ with respect to $t$ is $v^\prime=-te^{-\frac{t^{2}}{2}}$ (from Step 1). Then $\frac{d^{2}y}{dt^{2}}=(-1)e^{-\frac{t^{2}}{2}}+(-t)(-te^{-\frac{t^{2}}{2}})=(t^{2} - 1)e^{-\frac{t^{2}}{2}}$.

Step3: Determine where the second - derivative is negative

A function is concave down when $\frac{d^{2}y}{dt^{2}}<0$. Since $e^{-\frac{t^{2}}{2}}>0$ for all real $t$, we need to solve $t^{2}-1<0$. Factoring gives $(t - 1)(t + 1)<0$. The solution to this inequality is $-1

Answer:

$(-1,1)$