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I have a function defined as a composition of functions, one of which is the absolute value function. To give an equivalent expression for it, I have also defined it piecewise and have displayed it in a cases environment. The definition on each "piece" involves fractions, and the cases environment does not provide enough space. How do I increase the space? Is there a standard way to increase such inter-line spacing?

\documentclass[10pt]{amsart}


\begin{document}


\noindent \textbf{1.) }$f(x) = \sqrt{\vert 9 - 2x \vert}$. \\
\hspace*{2em}
\begin{tabular}{r l}
\textbf{i.)}    &   \hspace*{-0.5em}Is the the graph of $f$ symmetric across a line? \\
\textbf{ii.)}   &   \hspace*{-0.5em}Is $f$ continuous on $\mathbb{R}$? \\
\textbf{iii.)}  &   \hspace*{-0.5em}Is $f$ differentiable on $\mathbb{R}$? \\
\end{tabular}
\vskip0.2in


\noindent \textbf{Solution}
\begin{equation*}
f(x) =
\begin{cases}
\sqrt{2\left(\dfrac{9}{2} - x\right)}   &   \text{if } x \leq \dfrac{9}{2} \\
\sqrt{2\left(x - \dfrac{9}{2}\right)}   &   \text{if } x \geq \dfrac{9}{2}
\end{cases}
.
\end{equation*}
For every real number $t > 0$,
\begin{equation*}
f\left(\frac{9}{2} + t\right) = \sqrt{2t} = f\left(\frac{9}{2} - t\right) .
\end{equation*}
So, $f$ is symmetric across the vertical line $x = 9/2$.

\end{document}
Adelyn
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  • 1
    Just change the \\ to \\ \\ to add an extra blank line. Alternately, \\ takes an optional argument, such as \\[12pt] to add additional space. – Steven B. Segletes Aug 04 '16 at 17:05

1 Answers1

1

Use dcases from the mathtools package:

\documentclass[10pt]{amsart}
\usepackage{mathtools}

\begin{document}


\noindent \textbf{1.) }$f(x) = \sqrt{\vert 9 - 2x \vert}$. \\
\hspace*{2em}
\begin{tabular}{r@{ }l}
\textbf{i.)}    &  Is the the graph of $f$ symmetric across a line? \\
\textbf{ii.)}   &  Is $f$ continuous on $\mathbb{R}$? \\
\textbf{iii.)}  &  Is $f$ differentiable on $\mathbb{R}$? \\
\end{tabular}

\medskip

\noindent \textbf{Solution}
\begin{equation*}
f(x) =
\begin{dcases}
\sqrt{2\left(\frac{9}{2} - x\right)}   &   \text{if } x \leq \frac{9}{2}, \\
\sqrt{2\left(x - \frac{9}{2}\right)}   &   \text{if } x \geq \frac{9}{2}.
\end{dcases}
\end{equation*}
For every real number $t > 0$,
\begin{equation*}
f\left(\frac{9}{2} + t\right) = \sqrt{2t} = f\left(\frac{9}{2} - t\right) .
\end{equation*}
So, $f$ is symmetric across the vertical line $x = 9/2$.

\end{document}

Note also the other small adjustments I made.

enter image description here

egreg
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