determine the interval(s) on which the function (y(t)=e^{-\frac{t^{2}}{32}}) is concave up.

Explanation:

Step1: Find the first - derivative

Let $y(t)=e^{-\frac{t^{2}}{32}}$. Using the chain - rule, if $y = e^{u}$ and $u=-\frac{t^{2}}{32}$, then $\frac{dy}{du}=e^{u}$ and $\frac{du}{dt}=-\frac{t}{16}$. So, $y^\prime(t)=e^{-\frac{t^{2}}{32}}\left(-\frac{t}{16}\right)=-\frac{t}{16}e^{-\frac{t^{2}}{32}}$.

Step2: Find the second - derivative

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u =-\frac{t}{16}$ and $v = e^{-\frac{t^{2}}{32}}$. We have $u^\prime=-\frac{1}{16}$ and $v^\prime=e^{-\frac{t^{2}}{32}}\left(-\frac{t}{16}\right)$. Then $y^{\prime\prime}(t)=-\frac{1}{16}e^{-\frac{t^{2}}{32}}+\left(-\frac{t}{16}\right)\left(-\frac{t}{16}\right)e^{-\frac{t^{2}}{32}}=e^{-\frac{t^{2}}{32}}\left(\frac{t^{2}}{256}-\frac{1}{16}\right)$.

Step3: Determine where $y^{\prime\prime}(t)>0$

Set $y^{\prime\prime}(t)>0$. Since $e^{-\frac{t^{2}}{32}}>0$ for all real $t$, we only need to consider $\frac{t^{2}}{256}-\frac{1}{16}>0$. Let $x = t^{2}$, then $\frac{x}{256}-\frac{1}{16}>0$, or $x - 16>0$, so $x>16$. Substituting back $x = t^{2}$, we get $t^{2}>16$, which gives $t < - 4$ or $t>4$.

Answer:

$(-\infty,-4)\cup(4,\infty)$