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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)}
 \[