Use a computer algebra system to find the maximum value of $\left|f^{\prime \prime}(x)\right|$ on the closed interval. (This value is used in the error estimate for the Trapezoidal Rule, as discussed in Section 4.3.)
$f(x)=e^{-x^2 / 2}, \quad[0,1]$

Solution

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Step 1

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To find the $\textbf{maximum}$ of the given function we need to determine $\textbf{second derivative}$ of the given function.

$\textbf{Firsty}$ we will determine $\textbf{first}$ derivative of the function:

\begin{align*} f(x)&=e^{-x^2/2}\\ f'(x)&=-e^{-x^2/2}\cdot x\\ \end{align*}

$\textbf{Now}$ we can determine the second derivative.

To do that we will apply the $\textbf{product rule}$ :

\boxed{\color{#c34632}{[f(x)\cdot g(x)]'=f'(x)\cdot g(x)+f(x)\cdot g'(x)}}

$\textbf{This leads}$ to:

\begin{equation*} f''(x)=x^2e^{-x^2/2}-e^{-x^2/2}\\ \end{equation*}

To find solve the problem we need to find the $\textbf{maximum}$ of the:

\begin{equation} f''(x)=|x^2e^{-x^2/2}-e^{-x^2/2}|\\ \end{equation}

We will use $\textbf{graphing}$ utility.

So we can see that $\textbf{MAXIMUM}$ of the given function reaches for $x=0$ and its value is $f(0)=1$ .

So, our $\textbf{final result}$ is:

\boxed{\color{#c34632}{f(0)=1\Rightarrow MAXIMUM}}

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