Stitching Functions

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pieces and parts

Piecewise-Defined Functions

Piecewise defined functions are defined by using pieces of other functions.

Building Piecewise Defined Functions

Begin with several elementary functions with their own defining sets of domain, range, and pairs.
Select some pairs from each of those functions.
Collect these into a new set of pairs defining a new function.

Of course, we have to make sure that we do not duplicate domain numbers, otherwise we will not have a function. This new collection of pairs will automatically give us a domain by collecting all of the first coordinates.

For example, we could start with two constant functions: $Zero$ and $One$.

$Zero(r) = 0$ with $(-\infty , \infty )$ as its domain.
$One(r) = 1$ with $(-\infty , \infty )$ as its domain.

We can create a new function called step, by selecting some pairs from each of these functions.

From $Zero$ we’ll take pairs with negative domain values. This is one piece.
From $One$ we’ll take pairs with nonnegative domain values. This is another piece.

Note: Nonnegative means positive or 0.

Step Function

The step function uses different formulas depending on the domain number. If the domain number is negative, then the value of $step$ is $0$. If the value of the domain number is nonnegative, then the value of $step$ is $1$.

$step(-5.3) = 0$
$step(-0.7) = 0$
$step(0) = 0$
$step(1.2) = 1$
$step(142) = 1$

Graph of $y = step(x)$.

The step function uses two formulas, but only one at a time.

$step(x) = 0$, if $x < 0$
$step(x) = 1$, if $x \geq 0$

Either you use the formula $0$ or your use the formula $1$. The domain number at which you are evaluating $step$ tells you which formula to use.

The traditional way to write this formula looks like

\[ step(x) = \begin{cases} 0 & \text { if } x < 0 \\ 1 & \text { if } 0 \leq x \end{cases} \]

\[ step(x) = \begin{cases} 0 & (-\infty , 0) \\ 1 & [0, \infty ) \end{cases} \]

The formulas are listed in the left column and the domain conditions are listed in the right column. When evaluating a piecewise defined function, you don’t look for the formula first. First, you decide which domain condition in the right column your domain number satisfies. Then you choose the corresponding formula and evaluate with your domain number.

\[ G(t) = \begin{cases} 2t-1 & \text { if } t \leq -3 \\ t^2 & \text { if } t > -3 \end{cases} \]

$G(-5) = 2(-5) - 1 = -11$
$G(-3) = \answer {-7}$
$G(-2) = 2^2 = 4$
$G(0) = \answer {0}$
$G(3) = \answer {9}$

$-5$ and $-3$ satisfy $t \leq -3$, therefore we use $\answer {2t-1}$ as their formula.
$-2$ and $0$ and $3$ satisfy $t > -3$, therefore we use $\answer {t^2}$ as their formula.

We can have any number of pieces defining our function.

\[ p(k) = \begin{cases} k^2 + 1 & \text { if } -7 < k \leq -4 \\ -4k + 3 & \text { if } -4 < k \leq 1 \\ 1 - 3k & \text { if } k > 3 \end{cases} \]

\[ p(k) = \begin{cases} k^2 + 1 & (-7, -4] \\ -4k + 3 & (-4, 1] \\ 1 - 3k & (3, \infty ] \end{cases} \]

$p(-6) = \answer {37}$
$p(-2) = \answer {11}$
$p(-1) = \answer {7}$
$p(0) = \answer {3}$
$p(1) = \answer {-1}$
$p(2) = \answer {DNE}$
$p(3) = \answer {DNE}$
$p(4) = \answer {-11}$

From the formula above, we can see that the domain of $p$ is $(-7, 1] \cup (3, \infty )$.

This is because

\[ (-7, -4] \cup (-4, 1] \cup (3, \infty ) = (-7, 1] \cup (3, \infty ) \]

\[ T(v) = \begin{cases} 2v-1 & \text { if } -4 < v \leq -1 \\ -v+3 & \text { if } 1 \leq v < 7 \end{cases} \]

On the interval $(-4, -1]$, the graph should be a line segment for $T(v) = 2v-1$. On the interval $[1, 7)$, the graph should be another line segment for $T(v) = -v+3$

Graph of $y = T(v)$.

The line graph for $y = 2v - 1$ is only drawn over the domain interval $(-4, -1]$.
The line graph for $y = -v+3$ is only drawn over the domain interval $[1, 7)$.

The function $T$ has no minimum value. The maximum value of $T$ is $\answer {2}$, which occurs at $\answer {1}$.

Define the function $D(x)$ by the graph below of the equation $y = D(x)$.

The domain of $D$ is

$(-8,3) \cup (3,8)$ $[-8,3) \cup (3,8]$ $[-8,8]$

The range of $D$ is approximately

$[0.97, 8]$ $(-2.43, 8]$ $(-2.43, 6.38) \cup \{ 8 \}$ $(-2.43, 6.38] \cup \{ 8 \}$

The function $D$ has no minimum value

True False

The function $D$ has no maximum value

True False

The graph of a piecewise defined function may not look like pieces.

\[ f(x) = \begin{cases} \frac {1}{2} (x+8)^2 - \frac {21}{4} & \text { if } x \le -7 \\ x + \frac {9}{4} & \text { if } -7 \leq x \leq \frac {5}{2} \\ -(x-3)^2 + 5 & \text { if } \frac {5}{2} \le v \end{cases} \]

Graph of $y = f(x)$.

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more examples can be found by following this link
More Examples of Piecewise-Defined Functions

2025-01-07 02:29:41