# 4 types of relations (1 Viewer)

#### Jigmey

##### Member
Was away for the lesson so am unsure on how to do these questions. Also don’t know what the | x-3| means. Any help is appreciated.

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#### fan96

##### 617 pages
That's called the absolute value function, and it's defined by

$\bg_white |x| = \begin{cases} x, &\text{if }x \ge 0, \\-x, &\text{if }x < 0.\end{cases}$

Basically, you only want the magnitude of the number, not its sign.

For example,

$\bg_white |-1| = 1,$
$\bg_white |-2| = 2,$
$\bg_white |1| = 1,$
$\bg_white |0| = 0.$

If $\bg_white x = |y|$, then either $\bg_white x = y$ or $\bg_white x = -y$.

A useful identity to know is

$\bg_white |x| = \sqrt{x^2}.$

#### Jigmey

##### Member
That's called the absolute value function, and it's defined by

$\bg_white |x| = \begin{cases} x, &\text{if }x \ge 0, \\-x, &\text{if }x < 0.\end{cases}$

Basically, you only want the magnitude of the number, not its sign.

For example,

$\bg_white |-1| = 1,$
$\bg_white |-2| = 2,$
$\bg_white |1| = 1,$
$\bg_white |0| = 0.$

If $\bg_white x = |y|$, then either $\bg_white x = y$ or $\bg_white x = -y$.

A useful identity to know is

$\bg_white |x| = \sqrt{x^2}.$
so y = |x-3| would become y = x+3?

#### fan96

##### 617 pages
so y = |x-3| would become y = x+3?
Unfortunately it's not that simple. You can't "break up" the absolute value function like that.

If $\bg_white y = |x-3|$, then $\bg_white y = \pm(x - 3)$.

You can try graphing these to visualise the effect of the absolute value function.

#### TheShy

##### Member
You could go onto desmos and mess around with the absolute function to graphically gain an understanding of it

#### CM_Tutor

##### Well-Known Member
That's called the absolute value function, and it's defined by

$\bg_white |x| = \begin{cases} x, &\text{if }x \ge 0, \\-x, &\text{if }x < 0.\end{cases}$
So, expanding on the idea:

$\bg_white y = |x - 3| = \begin{cases} x - 3, &\text{if }x - 3 \geqslant 0 \implies x \geqslant 3, \\-(x - 3) = 3 - x, &\text{if }x - 3 < 0 \implies x < -3.\end{cases}$

FYI, the absolute value function is a special case of what is called the modulus (only appears in Extension 2) but which is also the basis for the use of the same vertical line symbols in vectors. $\bg_white |x|$ can be interpreted as the distance along the number line from the origin to x. Being a distance, it cannot be negative. Applying this idea, $\bg_white |x - 3|$ can be interpreted as the distance along the number line from 3 to x. This idea extends beyond one dimension, so that $\bg_white \overrightarrow{PQ}$ means the distance from P to Q and thus the length of the vector $\bg_white \overrightarrow{PQ}$.

#### CM_Tutor

##### Well-Known Member
Similar questions (so you can do the originals yourself):

1(a). Explain, with an example using a y-value, why each function is many-to-one.

(i) $\bg_white y = x^2 - 9$
(ii) $\bg_white y = |x - 2|$
(iii) $\bg_white y = (x - 2)x(x + 2)$
(iv) $\bg_white y = x^6 + 4$

(b) Hence classify each relation below.

(i) $\bg_white x = y^2 - 9$
(ii) $\bg_white x = |y - 2|$
(iii) $\bg_white x = (y - 2)y(y + 2)$
(iv) $\bg_white x = y^6 + 4$

1(a) If a graph passes the vertical line test - that is, that all values of x in the domain produce a single value of y, though the y-values can be different for each x-value - and thus the graph represents a function. That function is "many-to-one" if the graph fails the horizontal line test, and thus that there exists at least one value of y that leads to multiple (more than one) values of x. If the graph passes the horizontal line test then the function is "one-to-one." All four questions in 1(a) are functions as, in each case, choosing any value of x produces only one value of y.

(i) Consider $\bg_white y = 0$:

\bg_white \begin{align*} 0 &= x^2 - 9 \\ 9 &= x^2 \\ x &= \pm 3 \end{align*}

For this function, the value $\bg_white y = 0$ leads to two possible values of $\bg_white x \text{, } x = \pm 3$, and thus means the function is many-to-one instead of one-to-one.

(ii) Consider $\bg_white y = 1$:

\bg_white \begin{align*} 1 &= |x - 2| \\ \pm 1 &= x - 2 \\ x &= 2 \pm 1 = 1 \text{ or } 3 \end{align*}

For this function, the value $\bg_white y = 1$ leads to two possible values of $\bg_white x = 1 \text{ or } 3$, and thus means the function is many-to-one instead of one-to-one.

(iii) Consider $\bg_white y = 0$:

\bg_white \begin{align*} 0 &= (x - 2)x(x + 2) \\ 0 = x - 2 \quad \text{or} \quad 0 &= x \quad \text{or} \quad 0 = x + 2 \\ x &= 0 \text{ or } \pm 2 \end{align*}

For this function, the value $\bg_white y = 0$ leads to three possible values of $\bg_white x \text{, } x = -2, 0, \text{ or } 2$, and thus means the function is many-to-one instead of one-to-one.

(iv) Consider $\bg_white y = 68$:

\bg_white \begin{align*} 68 &= x^6 + 4 \\ 64 &= x^6 \\ \pm 8 &= x^3 \\ x &= \sqrt[3]{\pm 8} \\ x &= \pm 2 \end{align*}

For this function, the value $\bg_white y = 68$ leads to two possible values of $\bg_white x = \pm 2$, and thus means the function is many-to-one instead of one-to-one.

(b) Swapping x and y produces a relation that swaps the vertical and horizontal line results, so all of the relations in (b) will be one-to-many.

2(a). By solving for x, show that each function is one-to-one. (this method works only if x can be made the subject. Then, the relation is one-to-one if there is never more than one answer, and many-to-one otherwise.)

(i) $\bg_white y = 4 - 7x$
(ii) $\bg_white y = 27x^3$
(iii) $\bg_white y = \frac{2x}{x + 3}$

(b) Hence classify each relation below.

(i) $\bg_white x = 4 - 7y$
(ii) $\bg_white x = 27y^3$
(iii) $\bg_white x = \frac{2y}{y + 3}$

2(a)(i) and (b)(i):
\bg_white \begin{align*} y &= 4 - 7x \\ 7x &= 4 - y \\ x &= \frac{4 - y}{7} \end{align*}

Each value of x leads to one value of y, and each value of y leads to only one value of x. In other words, the graph passes both the horizontal and vertical line tests, and we have a one-to-one function - as must the relation in (b)(i).

2(a)(ii) and (b)(ii):
\bg_white \begin{align*} y &= 27x^3 \\ \frac{y}{27} &= x^3 \\ x &= \frac{\sqrt[3]{y}}{3} \end{align*}

Each value of x leads to one value of y, and each value of y leads to only one value of x. In other words, the graph passes both the horizontal and vertical line tests, and we have a one-to-one function - as must the relation in (b)(ii).

2(a)(iii) and (b)(iii):
\bg_white \begin{align*} y &= \frac{2x}{x + 3} \\ y(2x + 3) &= 2x \\ 2xy + 3y &= 2x \\ 3y &= 2x - 2xy = 2x(1 - y) \\ \frac{3y}{2(1 - y)} &= x \end{align*}

Each value of x leads to one value of y, and each value of y leads to only one value of x. In other words, the graph passes both the horizontal and vertical line tests, and we have a one-to-one function - as must the relation in (b)(iii).