Quiz:Piecewise definition of function - Revision history

Vipul: /* Converting to and from piecewise definitions */

2011-10-19T22:18:17Z

Converting to and from piecewise definitions

← Older revision		Revision as of 22:18, 19 October 2011
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	- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b - a)/2 \\ b - x, & (b - a)/2 \le x < b \\ x - b,& x \ge b \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b - a)/2 \\ b - x, & (b - a)/2 \le x < b \\ x - b,& x \ge b \\\end{array}\right.</math>
	+ <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b + a)/2 \\ b - x, & (b + a)/2 \le x < b \\ x - b, & x \ge b \\\end{array}\right.</math>		+ <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b + a)/2 \\ b - x, & (b + a)/2 \le x < b \\ x - b, & x \ge b \\\end{array}\right.</math>

			{Suppose <math>a_1 < a_2 < \dots < a_n</math> are <math>n</math> distinct real numbers. For an real number <math>x</math>, define <math>f(x)</math> as the minimum of the distances from <math>x</math> to each of the numbers <math>a_1,a_2,\dots,a_n</math>. In other words, <math>\! f(x) := \min \{ \|x - a_1\|, \|x - a_2\|, \|x - a_3\|, \dots, \|x - a_n\| \}</math>. How many different linear pieces does the correct piecewise linear definition of <math>f</math> have?
			\|type="()"}
			- <math>n</math>
			- <math>n + 1</math>
			- <math>2n - 2</math>
			- <math>2n - 1</math>
			+ <math>2n</math>
			\|\| The pieces will be <math>a_1 - x, x - a_1, a_2 - x,x - a_2, \dots, a_n - x, x - a_n</math>. There are <math>2n</math> pieces in total.

	</quiz>		</quiz>

Vipul: /* Converting to and from piecewise definitions */

2011-10-19T22:14:51Z

Converting to and from piecewise definitions

← Older revision		Revision as of 22:14, 19 October 2011
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	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>

	{Suppose <math>a < b</math> are real numbers. For an real number <math>x</math>, define <math>f(x)</math> as the minimum of the distances from <math>x</math> to <math>a</math> and <math>b</math>. In other words, <math>f(x) := \min \{ \|x - a\|, \|x - b\| \}</math>. Which of the following is the correct piecewise linear definition of <math>f</math>?		{Suppose <math>a < b</math> are real numbers. For an real number <math>x</math>, define <math>f(x)</math> as the minimum of the distances from <math>x</math> to <math>a</math> and <math>b</math>. In other words, <math>\! f(x) := \min \{ \|x - a\|, \|x - b\| \}</math>. Which of the following is the correct piecewise linear definition of <math>f</math>?
	\|type="()"}		\|type="()"}
	- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - (a + b)/2, & a <x < b \\ x - b, & x \ge b \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - (a + b)/2, & a <x < b \\ x - b, & x \ge b \\\end{array}\right.</math>

Vipul: /* Converting to and from piecewise definitions */

2011-10-19T22:14:00Z

Converting to and from piecewise definitions

← Older revision		Revision as of 22:14, 19 October 2011
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	{Suppose <math>a < b</math> are real numbers. For an real number <math>x</math>, define <math>f(x)</math> as the minimum of the distances from <math>x</math> to <math>a</math> and <math>b</math>. In other words, <math>f(x) := \min \{ \|x - a\|, \|x - b\| \}</math>. Which of the following is the correct piecewise linear definition of <math>f</math>?		{Suppose <math>a < b</math> are real numbers. For an real number <math>x</math>, define <math>f(x)</math> as the minimum of the distances from <math>x</math> to <math>a</math> and <math>b</math>. In other words, <math>f(x) := \min \{ \|x - a\|, \|x - b\| \}</math>. Which of the following is the correct piecewise linear definition of <math>f</math>?
	\|type="()"}		\|type="()"}
	- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - (a + b)/2, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - (a + b)/2, & a <x < b \\ x - b, & x \ge b \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - a, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - a, & a <x < b \\ x - b, & x \ge b \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\b - x, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\b - x, & a <x < b \\ x - b, & x \ge b \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b - a)/2 \\ b - x, & (b - a)/2 \le x < b \\ x - b, x \ge b \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b - a)/2 \\ b - x, & (b - a)/2 \le x < b \\ x - b,& x \ge b \\\end{array}\right.</math>
	+ <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b + a)/2 \\ b - x, & (b + a)/2 \le x < b \\ x - b, x \ge b \\\end{array}\right.</math>		+ <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b + a)/2 \\ b - x, & (b + a)/2 \le x < b \\ x - b, & x \ge b \\\end{array}\right.</math>

	</quiz>		</quiz>

Vipul: /* Converting to and from piecewise definitions */

2011-10-19T22:13:34Z

Converting to and from piecewise definitions

← Older revision		Revision as of 22:13, 19 October 2011
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	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>

	~~<quiz display=simple>~~
	{Suppose <math>a < b</math> are real numbers. For an real number <math>x</math>, define <math>f(x)</math> as the minimum of the distances from <math>x</math> to <math>a</math> and <math>b</math>. In other words, <math>f(x) := \min \{ \|x - a\|, \|x - b\| \}</math>. Which of the following is the correct piecewise linear definition of <math>f</math>?		{Suppose <math>a < b</math> are real numbers. For an real number <math>x</math>, define <math>f(x)</math> as the minimum of the distances from <math>x</math> to <math>a</math> and <math>b</math>. In other words, <math>f(x) := \min \{ \|x - a\|, \|x - b\| \}</math>. Which of the following is the correct piecewise linear definition of <math>f</math>?
			\|type="()"}
	- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - (a + b)/2, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - (a + b)/2, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - a, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - a, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>

Vipul at 22:11, 19 October 2011

2011-10-19T22:11:09Z

← Older revision		Revision as of 22:11, 19 October 2011
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	- <math>\left \lbrace\begin{array}{rl} 2x + 1, & x \ge -1/2 \\ -2x - 1, & x < -1/2 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 2x + 1, & x \ge -1/2 \\ -2x - 1, & x < -1/2 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>

			<quiz display=simple>
			{Suppose <math>a < b</math> are real numbers. For an real number <math>x</math>, define <math>f(x)</math> as the minimum of the distances from <math>x</math> to <math>a</math> and <math>b</math>. In other words, <math>f(x) := \min \{ \|x - a\|, \|x - b\| \}</math>. Which of the following is the correct piecewise linear definition of <math>f</math>?
			- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - (a + b)/2, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>
			- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\ x - a, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>
			- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\b - x, & a <x < b \\ x - b, x \ge b \\\end{array}\right.</math>
			- <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b - a)/2 \\ b - x, & (b - a)/2 \le x < b \\ x - b, x \ge b \\\end{array}\right.</math>
			+ <math>\left \lbrace\begin{array}{rl} a - x, & x \le a \\x - a, & a <x < (b + a)/2 \\ b - x, & (b + a)/2 \le x < b \\ x - b, x \ge b \\\end{array}\right.</math>

	</quiz>		</quiz>

Vipul at 21:57, 19 October 2011

2011-10-19T21:57:29Z

@@ Line 1: / Line 1: @@
 This quiz is related to [[piecewise definition of function]].
-==Pointwise combination==
+==Converting to and from piecewise definitions==
 <quiz display=simple>
 {Which of the following is the correct piecewise linear definition for <math>|x + 1| - |x|</math>?
 |type="()"}
@@ Line 20: / Line 12: @@
 - <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>
 </quiz>

Vipul: /* Pointwise combination */

2011-10-19T21:56:01Z

Pointwise combination

← Older revision		Revision as of 21:56, 19 October 2011
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	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 0, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 0, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>
	+ <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>		+ <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 2x + 1 < x \ge -1/2 \\ -2x - 1, & x < -1/2 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 2x + 1, & x \ge -1/2 \\ -2x - 1, & x < -1/2 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>

Vipul: /* Pointwise combination */

2011-10-19T21:55:47Z

Pointwise combination

← Older revision		Revision as of 21:55, 19 October 2011
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	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 0, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 0, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>
	+ <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>		+ <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 2x + 1 < x \ge -1/2 \\, -2x - 1, & x < -1/2 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 2x + 1 < x \ge -1/2 \\ -2x - 1, & x < -1/2 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>

Vipul: /* Pointwise combination */

2011-10-19T21:55:25Z

Pointwise combination

← Older revision		Revision as of 21:55, 19 October 2011
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	+ <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>		+ <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 2x + 1 < x \ge -1/2 \\, -2x - 1, & x < -1/2 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 2x + 1 < x \ge -1/2 \\, -2x - 1, & x < -1/2 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ ~~0, &~~ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>

	</quiz>		</quiz>

Vipul: /* Pointwise combination */

2011-10-19T21:54:54Z

Pointwise combination

← Older revision		Revision as of 21:54, 19 October 2011
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	\|type="()"}		\|type="()"}
	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge -1 \\ -1, & x < -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge -1 \\ -1, & x < -1 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 0, & -1 < x < 0 \\, -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 0, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>
	+ <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>		+ <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 2x + 1, & -1 < x < 0 \\ -1, & x \le -1 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 2x + 1 < x \ge -1/2 \\, -2x - 1, & x < -1/2 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 2x + 1 < x \ge -1/2 \\, -2x - 1, & x < -1/2 \\\end{array}\right.</math>
	- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 0, & 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>		- <math>\left \lbrace\begin{array}{rl} 1, & x \ge 0 \\ 0, & 2x + 1, & -1/2 < x < 0 \\ -2x - 1, & -1 \le x \le -1/2 \\ -1, & x \le -1 \\\end{array}\right.</math>