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Given the following function,
$$f(x) = \left\{ \begin{array}{ccc}
x^2-3x & \textrm{ if } & x \lt -1\\\\
2x-3 & \textrm{ if } & -1 \le x \lt 1\\\\
2 & \textrm{ if } & x = 1\\\\
\displaystyle{\frac{1}{x-2}} & \textrm{ if } & x \ge 1\\\\
\end{array} \right.$$
Evaluate the following expressions. When a limit below fails to exist, cite the reason why this is the case.
$f\,(-1)$
$f\,(1)$
$\displaystyle{\lim_{x \rightarrow -1} \ f\,(x)}$
$\displaystyle{\lim_{x \rightarrow 0} \ f\,(x)}$
$\displaystyle{\lim_{x \rightarrow 1} \ f\,(x)}$
$\displaystyle{\lim_{x \rightarrow 2} \ f\,(x)}$
Now identify all discontinuities in the above function, giving a graphical interpretation (with justification) of each.
For each given function, identify all of the discontinuities, classify each as removable or non-removable, and provide a graphical interpretation (hole, gap, vertical asymptote, etc...). Write your answer in paragraph form, and make sure to justify your claims with appropriate limiting values, functional values, and the definition of continuity.
$H(x) = \displaystyle{\frac{x^2-4}{x-2}}$
$F(x) = \displaystyle{\frac{x+3}{x^2-9}}$
$f(x) = \displaystyle{\frac{|x+4|}{x}}$
$\displaystyle{g(x) = \left\{ \begin{array}{ccc}
x^3 & \textrm{ if } & x
0 & \textrm{ if } & x=0\\
x^2 & \textrm{ if } & x>0
\end{array} \right.}$
$h(x)=\tan 2x$ for $[0,\pi]$
$G(x)=\sqrt{x-3}$
$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
\sin 2x&,& x \lt -\frac{\pi}{2}\\\\
\cos x &,& -\frac{\pi}{2} \lt x \le 0\\\\
x+1 &,& 0 \lt x \lt 2\\\\
\sqrt{x+2} &,& 2 \le x \lt 7\\\\
\frac{1-x}{x-9} &,& x \gt 7\\\\
\end{array}\right.}$
For each given function, identify all of the discontinuities, classify each as removable or non-removable, and provide a graphical interpretation (hole, gap, vertical asymptote, etc...). Write your answer in paragraph form, and make sure to justify your claims with appropriate limiting values, functional values, and the definition of continuity.
$\displaystyle{f(x) = \frac{x+3}{x^2-9}}$
$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
\frac{2x+3}{x+2} &,& x \lt 1\\\\
x^2 &,& 1 \le x \lt 2\\\\
5x-6 &,& x \ge 2\\
\end{array}\right.}$
$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
e^{x^2} &,& x \lt 0\\\\
\sqrt{1-x^2} &,& 0 \lt x \le 1\\\\
-2 &,& 1 \lt x \lt \pi\\\\
2\sec x &,& \pi \le x \lt 2\pi\\\\
2 &,& x \gt 2\pi\\\\
\end{array}\right.}$
$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
\sin x &,& x \lt -3\pi/2\\\\
\tan(x/2) &,& -3\pi/2 \lt x \le 0\\\\
x^3+x &,& 0 \lt x \le 1\\\\
\frac{-3x+1}{x-2} &,& 1 \lt x \lt 3\\\\
-\sqrt{x+6} &,& x \ge 3\\\\
\end{array}\right.}$
$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
\frac{x}{4+x} &,& x \lt -2\\\\
2x+1 &,& -2 \le x \lt 0\\\\
1+\sqrt{x} &,& 0 \lt x \lt 4\\\\
2 &,& x = 4\\\\
x^2-3x-1 &,& x \gt 4\\
\end{array}\right.}$
$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
\frac{x+5}{x+6} &,& x \le -5\\\\
\sqrt{25-x^2} &,& -5 \lt x \le 0\\\\
\tan 2x &,& 0 \lt x \lt \pi\\\\
\sin x &,& \pi \le x \lt 2\pi\\\\
\frac{x}{2} - \pi &,& x \gt 2\pi
\end{array}\right.}$
$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
(x+6)^{1/3} - 1 &,& x \lt -5 \\\\
\sqrt{25-x^2} &,& -5 \le x \lt -3 \\\\
e^x - x &,& -3 \lt x \le 0 \\\\
\cos x &,& 0 \lt x \lt \frac{\pi}{2} \\\\
\sin 3x &,& \frac{\pi}{2} \le x \lt \pi \\\\
2|x-5| &,& \pi \le x \lt 6
\end{array} \right.}$
$\displaystyle{g(x) = \left\{ \begin{array}{ccc}
\frac{-x}{x+4} &,& x \le -2\\\\
x+5 &,& -2 \lt x \lt 0\\\\
\sqrt{25-x^2} &,& 0 \lt x \le 4\\\\
(31-x)^{1/3} &,& x \gt 4
\end{array} \right.}$
Find the value of $k$ that will make the function below continuous.
$$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
\frac{\sqrt{x+3}-3}{x^2-36} &,& 0 \le x \lt 6\\\\
k &,& x \ge 6\\
\end{array}\right.}$$
Are there values of $a$ and $b$ that will make the function below continuous? If there are, find them. If not, explain why.
$$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
\frac{x-4}{x^2-16} &,& x \neq 4,-4\\\\
a &,& x=4\\\\
b &,& x=-4\\\\
\end{array}\right.}$$
Given the following piecewise defined function, g(x)
$$\displaystyle{g(x) = \left\{ \begin{array}{ccc}
\frac{x^2+x-6}{x^2-4x+4} & \textrm{ if } & x \ne 2\\\\
k & \textrm{ if } & x=2\\
\end{array} \right.}$$
Is there a value for $k$ that makes $g(x)$ continuous everywhere? Clearly explain your reasoning, finding this value of $k$ if it exists.
Given the following function
$$f(x) = \left\{ \begin{array}{ccc}
x-3 & \textrm{ if } & x \lt 1\\\\
cx^2+1 & \textrm{ if } & x \ge 1\\\\
\end{array} \right.$$
Determine if there is a value for $c$ such that the function is continuous at $x=1$. Clearly state your reasoning.
If possible, find values for $m$ and $n$ to make the following function continuous everywhere. If this is not possible, explain clearly why not.
$$f(x) = \left\{ \begin{array}{ccc}
\frac{x^3-27}{x^2-9} &,& x \neq \pm 3\\\\
m &,& x = -3\\\\
n &,& x = 3
\end{array} \right.$$
Given the following function
$$f(x) = \left\{ \begin{array}{ccc}
-\sqrt{36-x^2} & \textrm{ if } & x \lt 0\\\\
2x & \textrm{ if } & 0 \lt x \lt 2\\\\
\displaystyle{\frac{-2x^2}{x-4}} & \textrm{ if } & x \ge 2
\end{array} \right.$$
Discuss any discontinuities present in this function. Determine if each is removable or nonremovable, and describe what behavior is seen in the graph of $y=f(x)$ at these points.