Piecewise-defined functions are functions where the formula used depends upon the value of the input. When looking for discontinuities on piecewise-defined functions, you need to investigate the behavior at values where the formula changes as well as values where the issues discussed in Activity 2.12 might pop up.

# Subsection2.13.1Exercises

This question is all about the function $f$ defined by

\begin{equation*} \fe{f}{x}=\begin{cases}\frac{4}{5-x}&x\lt1\\\frac{x-3}{x-3}&1\lt x\lt4\\2x+1&4\leq x\leq7\\\frac{15}{8-x}&x\gt7\text{.}\end{cases} \end{equation*}
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Complete Table 2.13.1.

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Complete Table 2.13.2.

Consider the function $f$ defined by $\fe{f}{x}=\begin{cases}\frac{5}{x-10}&x\leq5\\\frac{5}{5x-30}&5\lt x\lt7\\\frac{x-2}{12-x}&x>7\text{.}\end{cases}$

State the values of $x$ where each of the following occur. If a stated property doesn't occur, make sure that you state that (as opposed to simply not responding to the question). No explanation necessary.

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At what values of $x$ is $f$ discontinuous?

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At what values of $x$ is $f$ continuous from the left, but not from the right?

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At what values of $x$ is $f$ continuous from the right, but not from the left?

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At what values of $x$ does $f$ have removable discontinuities?

Consider the function $g$ defined by $\fe{g}{x}=\begin{cases}\frac{C}{x-17}&x\lt10\\C+3x&x=10\\2C-4&x\gt10\text{.}\end{cases}$

The symbol $C$ represents the same real number in each of the piecewise formulas.

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Find the value for $C$ that makes the function continuous on $\ocinterval{-\infty}{10}\text{.}$ Make sure that your reasoning is clear.

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Is it possible to find a value for $C$ that makes the function continuous over $\ointerval{-\infty}{\infty}\text{?}$ Explain.