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Determine where the derivatives of the following functions fail to exist, characterizing the behavior of the related graph (i.e., corner, cusp, vertical tangent, hole, gap, vertical asymptote, etc...) at each such $x$-value. Classify any discontinuities found as either removable or non-removable. Justify your claims with appropriate function values, derivative values, and limiting values.
$\displaystyle{f\,(x) = |x+2|}$
$f(x) = (x+1)^{2/3}$
$g(x) = 1 - \sqrt[3]{x}$
$\displaystyle{f\,(x) = \left\{ \begin{array}{rlc}
x^2+1 &,& x \lt 1\\\\
3 &,& x = 1\\\\
2 \sqrt{2x-1} &,& x \gt 1
\end{array} \right.}$
$\displaystyle{g(x) = \left\{ \begin{array}{rlc}
x+1 &,& x \le 2\\\\
1 &,& 2 \lt x \le 4\\\\
x^2 - 8x + 17 &,& x \gt 4
\end{array} \right.}$
$\displaystyle{h(x) = \left\{ \begin{array}{rlc}
2-x &,& x \le 3\\\\
x^2 - 10 &,& x \gt 3
\end{array} \right.}$
$\displaystyle{f\,(x) = \left\{ \begin{array}{rlc}
\sqrt{16-x^2} &,& x \lt 0\\\\
x^3+4 &,& 0 \le x \le 1\\\\
-x^2+5 &,& x \gt 1
\end{array} \right.}$
$\displaystyle{f\,(x) = \left\{ \begin{array}{rlc}
\frac{9}{x^2-4} &,& x < -1\\\\
3\ln(-x)-3 &,& -1 \le x \le 0\\\\
\frac{2x^2-x}{2x-1} &,& 0 < x < 1\\\\
\sin(x-1)+1 &,& 1 < x \le \pi + 1\\\\
2(x-5)^{2/3} &,& \pi + 1 < x \le 6\\\\
5-\sqrt{-x^2+4x+21} &,& 6 < x < 7\\\\
\sqrt[3]{x-7}+5 &,& x \ge 7
\end{array} \right.}$
$\displaystyle{f\,(x) = \left\{ \begin{array}{rlc}
x^2-5 &,& x \le -2\\\\
x+1 &,& -2 \lt x \lt 0\\\\
-1 &,& x = 0\\\\
\sqrt{4-x^2} - 1 &,& 0 \lt x \lt 2\\\\
2 &,& x \ge 2
\end{array} \right.}$
$\displaystyle{h(x) = \left\{ \begin{array}{ccc}
x+9 &,& x \le -9\\\\
(x+1)^{1/3} &,& -9 \lt x \lt 0\\\\
\cos x &,& 0 \le x \lt \pi\\\\
\displaystyle{-\tan \left( \frac{x}{4} \right)} &,& \pi \lt x \le 3\pi\\\\
\displaystyle{\frac{1}{x-12}} &,& x \gt 3\pi
\end{array} \right.}$
$\displaystyle{g(x) = \left\{ \begin{array}{ccc}
|x+7| &,& x \le -4\\\\
\sqrt{25-x^2} &,& -4 \lt x \lt 0\\\\
5-x^2 &,& 0 \le x \lt 2\\\\
\frac{-x}{x-4} &,& x \gt 2
\end{array} \right.}$
$\displaystyle{f\,(x) = \left\{ \begin{array}{ccc}
x^3-3x &,& x \lt 1\\\\
-2 &,& x = 1\\\\
x^2 + 1 &,& x \gt 1
\end{array} \right.}$
$\displaystyle{g(x) = \left\{ \begin{array}{ccc}
\frac{-x}{x+4} &,& x \le -2\\\\
x+5 &,& -2 < x < 0\\\\
\sqrt{25-x^2} &,& 0 < x \le 4\\\\
(31-x)^{1/3} &,& x > 4
\end{array} \right.}$
$\displaystyle{g(x) = \left\{ \begin{array}{ccc}
x^2 - 1 &,& x \gt 3\\\\
6x - 10 &,& x \le 3
\end{array} \right.}$
$\displaystyle{g(x) = \left\{ \begin{array}{ccc}
x/2 &,& x \le -2\\\\
x^2 &,& -2 \lt x \lt 1\\\\
-x+5 &,& x \ge 1
\end{array} \right.}$
$\displaystyle{f(x) = \left\{ \begin{array}{ccc}
(x+6)^{1/3} - 1 & \textrm{ if } & x < -5 \\\\
\sqrt{25-x^2} & \textrm{ if } & -5 \le x < -3 \\\\
e^x - x & \textrm{ if } & -3 < x \le 0 \\\\
\cos x & \textrm{ if } & 0< x < \frac{\pi}{2} \\\\
\sin 3x & \textrm{ if } & \frac{\pi}{2} \le x < \pi \\\\
2|x-5| & \textrm{ if } & \pi \le x < 6
\end{array} \right.}$
$\displaystyle{f\,(x) = \left\{ \begin{array}{ccc}
-x -1 &,& x \lt -1\\\\
x^2-1 &,& -1 \le x \le 1\\\\
2x-2 &,& x \gt 1
\end{array} \right.}$
$\displaystyle{f\,(x) = \left\{ \begin{array}{ccc}
5-x^2 &,& x \lt 0\\\\
3 &,& x=0\\\\
\sqrt{25-x^2} &,& 0 \lt x \lt 4\\\\
1-4(x-5)^{1/3} &,& 4 \le x \lt 6\\\\
\frac{3-3x}{2x-7} &,& x \le 6
\end{array} \right.}$