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midterm should be done

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Claudio Maggioni 2020-04-03 22:27:28 +02:00
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@ -91,6 +91,20 @@ $$ \left|f'(x) - \frac{f(x + h) - f(x - h)}{2h}\right| \leq \left|f'(x) - \left(
\Rightarrow
C = \frac{f'''(x)}{6}$$
\subsection*{Point b)}
In order to find a valid constant for the entire domain $[-10,10]$ we find the constant for the value of $x$ that maximizes $f'''(x)$, hence the
$\sup$s.
\paragraph{Function 1}
$$f(x) = e^{-x^2} \hspace{1cm} f'''(x) = -4e^{-x^2} (2x^3 - 3x) \hspace{1cm} C = \sup_{x \in [-10,10]} \frac{f'''(x)}{6} = 2 \sqrt{1-\sqrt{\frac{2}{3}}}e^{\sqrt{\frac{3}{2}} - \frac{3}{2}} $$
\paragraph{Function 2}
$$f(x) = x^2 \hspace{1cm} f'''(x) = 0 \hspace{1cm} C = \sup_{x \in [-10,10]} \frac{f'''(x)}{6} = 0 $$
\paragraph{Function 3}
$$f(x) = sin(x) \hspace{1cm} f'''(x) = -cos(x) \hspace{1cm} C = \sup_{x \in [-10,10]} \frac{f'''(x)}{6} = \frac{1}{6} $$
\section*{Question 4}
\subsection*{Point a)}
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