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Use a computer algebra system to find the maximum value of f(x)\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)=ex2/2,[0,1]f(x)=e^{-x^2 / 2}, \quad[0,1]


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

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

f(x)=ex2/2f(x)=ex2/2x\begin{align*} f(x)&=e^{-x^2/2}\\ f'(x)&=-e^{-x^2/2}\cdot x\\ \end{align*}

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

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

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

This leads\textbf{This leads} to:

f(x)=x2ex2/2ex2/2\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 maximum\textbf{maximum} of the:

f(x)=x2ex2/2ex2/2\begin{equation} f''(x)=|x^2e^{-x^2/2}-e^{-x^2/2}|\\ \end{equation}

We will use graphing\textbf{graphing} utility.

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

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

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

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