wip midterm
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0 \\ 0 \\ 0 \\ 0 \\ 0 \\
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0 \\ 0 \\ 0 \\ 0 \\ 0 \\
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\end{bmatrix}
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\end{bmatrix}
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\]
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\section*{Question 5}
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\subsection*{Point a)}
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$$f(x) = x \hspace{2cm} K_{abs} = |f'(x)| = 1 \hspace{2cm} K_{rel} = \left|\frac{1 \cdot x}{x}\right| = 1$$
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\subsection*{Point b)}
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$$f(x) = \sqrt[3]{x} \hspace{2cm} K_{abs} = |f'(x)| = \frac{1}{3\sqrt[3]{x^2}} \hspace{2cm}
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K_{rel} = \left|\frac{1}{3\sqrt[3]{x^2}} \cdot \frac{x}{\sqrt[3]{x}}\right| = \frac{1}{3}$$
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\subsection*{Point c)}
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$$f(x) = \frac{1}{x} \hspace{2cm} K_{abs} = |f'(x)| = \frac{1}{x^2} \hspace{2cm}
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K_{rel} = \left|\frac{-x}{x^2} \cdot \frac{1}{\frac{1}{x}}\right| = 1$$
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\subsection*{Point d)}
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$$f(x) = e^x \hspace{2cm} K_{abs} = |f'(x)| = e^x \hspace{2cm}
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K_{rel} = \left|\frac{xe^x}{e^x}\right| = |x|$$
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\subsection*{Point e)}
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Cases \textit{a)},\textit{b)} and \textit{c)} are well-conditioned for any $x$ since their $K_rel$
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is not defined by x. Case \textit{d)} is well-conditioned only for $x$s whose absolute value is in the order of magnitude of $1$ or less, since $K_rel$ in this case is exactly $|x|$.
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\end{document}
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\end{document}
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