Quiz:Piecewise definition of function: Difference between revisions
| (2 intermediate revisions by the same user not shown) | |||
| Line 12: | 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> | - <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> | ||
- <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> | ||
{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> | ||
Latest revision as of 22:18, 19 October 2011
This quiz is related to piecewise definition of function.
Converting to and from piecewise definitions
Pointwise combination (computational)
Continuity and pointwise combination
Composition
Fill this in later
Differentiation
Fill this in later