Shifting Domain

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

same characteristics

One way to create a new function from an existing function is to shift all of the domain numbers in the function pairs.

$(a, b) \mapsto (a + k, b)$

$\blacktriangleright$ How do we describe a shifting domain?

We want a graphical description. We want an algebraic description.

Below is the piecewise-defined function, $T(v)$ . The variable $v$ is representing the domain values from the set $(-4,-1] \cup [1,7)$ .

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

On the interval $(-4, -1]$ , the graph is a line segment for $T(v) = 2v-1$ , a linear function with a restricted domain.

On the interval $[1, 7)$ , the graph is another line segment for $T(v) = -v+3$ , a linear function with a restricted domain.

Graph of $y = T(v)$ .

The domain of $T$ has two maximal intervals:, $(-4,-1]$ and $[1,7)$ . These correspond to two line segments on the graph. The endpoints give us four strategic points on the graph:

$(-4, -9)$ , which is an open dot on the graph.
$(-1, -3)$ , which is a closed dot on the graph.
$(1, 2)$ , which is a closed dot on the graph.
$(7, -4)$ , which is an open dot on the graph.

A New Function

$B(k)$ is a new function we are creating, but its definition is based on $T(v)$ .

$\blacktriangleright$ Define $B$ by $B(k) = T(k+3)$ with the induced domain.

First Question: What is the domain of $B$ ?

$k$ is representing the values of the domain of $B$ and $k+3$ is now representing the domain values of $T$ . Therefore $k+3$ must take on the values in $(-4,-1] \cup [1,7)$ .

$k+3 \in (-4,-1] \cup [1,7)$

$k$ must be in a similar set, but shifted to the left by $3$ , because we need to subtract $3$ from $k+3$ to get $k$ .

$k \in \left ( -7,\answer {-4} \right ] \cup \left [ \answer {-2},4 \right )$

The domain of $B$ is $(-7,-4] \cup [-2,4)$ . It was induced from the domain of $T$ through the definition of $B$ , which is using $T$ . The values in the domain of $B$ are real numbers, $k$ , such that $k+3$ is in the domain of $T$ .

The domain has shifted.

But, the structure of the function hasn’t changed. The same formulas are in use. They are just used by different domain numbers.

Second Question: The graph of $T$ had four strategic points. What are the corresponding points on the graph of $B$ ?

We have four strategic points on the graph of $T$ :

$(-4, -9)$ , which is an open dot on the graph of $T$ .
$(-1, -3)$ , which is a closed dot on the graph of $T$ .
$(1, 2)$ , which is a closed dot on the graph of $T$ .
$(7, -4)$ , which is an open dot on the graph of $T$ .

How do these translate to strategic points on the graph of $B$ ?

$(-4, -9)$ tells us that $T(-4) = -9$ . To match the template $B(k) = T(k+3)$ , we need $k=-7$ . Replacing $k$ with $-7$ gives us $B(-7) = T(-7 + 3) = T(-4) = -9$ $B(-7) = -9$ and the point $(-7, -9)$ is on the graph of $B$ . It is an open dot.
$(-1, -3)$ tells us that $T(-1) = -3$ . To match the template $B(k) = T(k+3)$ , we need $k=-4$ . Replacing $k$ with $-4$ gives us $B(-4) = T(-4 + 3) = T(-1) = -3$ $B(-4) = -3$ and the point $(-4, -3)$ is on the graph of $B$ . It is a closed dot.
$(1, 2)$ tells us that $T(1) = 2$ . To match the template $B(k) = T(k+3)$ , we need $k=-2$ . Replacing $k$ with $-2$ gives us $B(-2) = T(-2 + 3) = T(1) = 2$ $B(-2) = 2$ and the point $(-2, 2)$ is on the graph of $B$ . It is a closed dot.
$(7, -4)$ tells us that $T(7) = -4$ . To match the template $B(k) = T(k+3)$ , we need $k=4$ . Replacing $k$ with $4$ gives us $B(4) = T(4 + 3) = T(7) = -4$ $B(4) = -4$ and the point $(4, -4)$ is on the graph of $B$ . It is an open dot.

Third Question: What is the formula for $B$ ?

When $k \in (-7,-4]$ , then $k+3 \in \left (\answer {-4},\answer {-1}\right ]$ .

$v=k+3$ , so $v \in \left (\answer {-4},\answer {-1}\right ]$ and we are using the formula $2v-1$ with $v=k+3$ .

All together: when $k \in (-7,-4]$ , then $B(k) = 2(k+3)-1$ .

When $k \in [-2,4)$ , then $k+3 \in \left [\answer {1},\answer {7}\right )$ .

$v=k+3$ , so $v \in \left [\answer {1},\answer {7}\right )$ and we are using the formula $-v+3$ with $v=k+3$ .

All together: when $k \in [-2,4)$ , then $B(k) = -(k+3)+3$ .

$B(k) = \begin {cases} 2k+5 & \text { if } -7 < k \leq -4 \\ -k & \text { if } -2 \leq k < 4 \end {cases}$

Graph of $z = B(k)$ .

When we look at the graphs side-by-side, we can see that the graph of $T$ has simply shifted left $3$ to become the graph of $B$ .

The structure of the two graphs hasn’t changed.

From the definition, $B(k) = T(k+3)$ , we can see that $v=k+3$ , or $k=v-3$ . All of the $k$ -values are $3$ less than the $v$ -values. $B$ is $T$ shifted left $3$ .

Variables

Notice that four different variables, $v$ , $y$ , $k$ , and $z$ , are used to describe the domain and range values of $T$ and $B$ . We are using separate names for the different types of values to keep everything clear. If we use the same variable to represent both domains, then we run into problems.

For instance... Suppose we use $x$ to represent values in both domains. Then $x = 0$ is a possibility for $B$ . $B(0) = 0$ . However, if we write $x = 0$ and $x$ is also representing the domain of $T$ , then we are implying that $T(0)$ is a valid number. However, $0$ is not in the domain of $T$ .

We use variables to help us think about structure and relationships. For this reason, we should not use the same variable to represent values from two different sets in the same situation.

Similarly, the two vertical axes have been labelled $y$ and $z$ . If we use $y$ to represent both axes, then we are using $y$ to represent both ranges and that also leads to communication problems.

We will avoid using the same variables or labels for values from different sets.

Shifting the Domain

$V(h) = \begin {cases} -2h-3 & \text { on } [-6, -2] \\ -(h+3)(h-3) & \text { on } (-2, 4] \\ \frac {7h}{4} - 8 & \text { on } (4,6) \end {cases}$

A graph always helps our thinking. Here is the graph of $y = V(h)$ .

Define a new function by $f(x) = V(x-1)$ with the induced domain.

The domain of $V$ is $[-6, 6)$ . These values are represented by $x-1$ in this definition, because $x-1$ is the expression inside the parentheses for $V$ and that is where the expression lives for the domain values of $V$ .

$x$ represents the domain values of $f$ . If $x - 1 \in [-6, 6)$ , then $x \in \left [\answer {-5}, \answer {7}\right )$ . This is the domain for $f$ . It is shifted $1$ to the right from the domain of $V$ . The individual maximal intervals become $[-5, -1]$ , $\left (\answer {-1}, \answer {5}\right ]$ , and $[5, 7)$ .

$f(x) = V(x-1) = \begin {cases} -2(x-1)-3 & \text { on } [-5, -1] \\ -((x-1)+3)((x-1)-3) & \text { on } (-1, 5] \\ \frac {7(x-1)}{4} - 8 & \text { on } (5, 7) \end {cases}$

$f(x) = \begin {cases} -2x - 1 & [-5, -1] \\ -(x+2)(x-4) & (-1, 5] \\ \frac {7(x-1)}{4} - 8 & (5, 7) \end {cases}$

The graph has the same three shapes. The corresponding endpoints are hollow or filled as they were in the graph of $y=V(h)$ .

Shifting Domain

Let $P(t)$ be a function with domain $[-11, -7) \cup \{ -6 \} \cup [-4,3) \cup (4, 7] \cup \{ 8 \} \cup [9, 10)$ .

Let $g(w)$ be defined as a shifted domain version of $P(t)$ . The domain of $g(w)$ is $[-7, -3) \cup \{ -2 \} \cup [0,7) \cup (8, 11] \cup \{ 12 \} \cup [13, 14)$ .

In terms of $t$ , $w = \answer {t+4}$
In terms of $w$ , $t = \answer {w-4}$ .

$g(w) = P\left (\answer {w-4}\right )$
$P(t) = g\left (\answer {t+4}\right )$

When a new function is defined as a domain shift of an existing function, then there are two different domains.

It is natural to think, ”What was done to the old domain to get the new one?”

Suppose $f(t)$ is an existing function with domain $[-3, 5]$ .

Define a new function $g(x)$ as $g(x) = f(x-8)$ with the induced domain.

“ $t$ ” represents elements of the old domain of $f(t)$ .
“ $x$ ” represents elements of the new domain of $g(x)$ .

It is natural to ask “what happens to $t$ to get $x$ ?”

But that is NOT how it is presented to us.

We are told that $t=x-8$ . But that is what happens to $x$ to get $t$ . That is backwards of how we naturally think about old and new. We need to solve for $x$ to see what is done to $t$ .

$t+8=x$

To get the new domain number, $x$ , add $8$ to the old domain numbers $t$ . This tells us that the graph of $g$ will be shifted to the right by $8$ , compared to the position of the graph of $f$ .

It is reversed from how it was originally defined, $g(x) = f(x-8)$ , because the “inside” of $f$ is $x-8$ here. That is not what is done to $t$ . When we solve $t=x-8$ for $x$ , we reverse all of the arithmetic and discover what was done to $t$ .

Shifting Domain

Let $D(y)$ be a function with its domain.

Let $K(r)$ be defined as $K(r) = D(r+7)$ with its induced domain.

Then the domain of $K$ is the domain of $D$ , but shifted Left Right by 5 6 7

Then the graph of $K$ is the graph of $D$ , but shifted Left Right by 5 6 7

Shifting Domain

Let $f(w)$ be a function with its domain.

Let $g(t)$ be defined as $g(t) = f(t-3)$ with its induced domain.

Then the domain of $g$ is the domain of $f$ , but shifted Left Right by 1 2 3

Then the graph of $g$ is the graph of $f$ , but shifted Left Right by 1 2 3

ooooo=-=-=-=-=-=-=-=-=-=-=-=-=ooOoo=-=-=-=-=-=-=-=-=-=-=-=-=ooooo
more examples can be found by following this link
More Examples of Shifting