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# Which of the following rules represents a function

## Tags: Mathematik     RASTA 17:14 Uhr, 08.01.2022

Tag,
Which of the following rules represents a function?
Welche der folgenden Regeln stellt eine Funktion dar?

can you explain why these are called rules?.
$i\phantom{\rule{0.12em}{0ex}}$ know about functions but confused here about they call these rules?
please, any hinto with the subject/topic at hand. can you refer a tutorial or begin a solution. however way you think it is best, now you know $i\phantom{\rule{0.12em}{0ex}}$ am here to get you to do my homework but to explain the way I should go and the materials available. you can refer german sites too. $i\phantom{\rule{0.12em}{0ex}}$ dont speak much but can read German.

Danke im Voraus.

Which of the following rules represents a function?

$\left\{\left(x\phantom{\rule{0.12em}{0ex}},y\phantom{\rule{0.12em}{0ex}}\right):y\phantom{\rule{0.12em}{0ex}}2=x\phantom{\rule{0.12em}{0ex}}2\right\}$
$\left\{\left(x\phantom{\rule{0.12em}{0ex}},y\phantom{\rule{0.12em}{0ex}}\right):y\phantom{\rule{0.12em}{0ex}}=x\phantom{\rule{0.12em}{0ex}}2\right\}$
$\left\{\left(x\phantom{\rule{0.12em}{0ex}},y\phantom{\rule{0.12em}{0ex}}\right):|y\phantom{\rule{0.12em}{0ex}}|=x\phantom{\rule{0.12em}{0ex}}2\right\}$
$\left\{\left(x\phantom{\rule{0.12em}{0ex}},y\phantom{\rule{0.12em}{0ex}}\right):y\phantom{\rule{0.12em}{0ex}}2=x\phantom{\rule{0.12em}{0ex}}\right\}$  supporter 17:38 Uhr, 08.01.2022

123mathe.de/relationen-und-funktionen  RASTA 18:04 Uhr, 08.01.2022

thank you. Ich melde mich noch!. A "function rule" simply is an equation which defines a function - it describes a functional relationship.
$y\phantom{\rule{0.12em}{0ex}}={x\phantom{\rule{0.12em}{0ex}}}^{2}+1$ is an example for a function rule.
${y\phantom{\rule{0.12em}{0ex}}}^{4}=x\phantom{\rule{0.12em}{0ex}}$ is a rule, but its not a function rule, because a function has to be unique (every $x\phantom{\rule{0.12em}{0ex}}$ is assigned just one $y\phantom{\rule{0.12em}{0ex}}\right)$. Here for $x\phantom{\rule{0.12em}{0ex}}=1$ we have two possible values for $y\phantom{\rule{0.12em}{0ex}},$ either $+1$ or $-1$. So this rules defines a relation which is not unique and therefore is not a function.

Now check your given rules if they define a function (unique relation) or not.  RASTA 21:07 Uhr, 08.01.2022

I do not understand well,
can you provide an example?.
Graphing the equations given in Desmos I could see that $b\phantom{\rule{0.12em}{0ex}}\right)$ is the right choice becasue it passes the vertical test,
but, they won't let me do it with the graph,
how can $i\phantom{\rule{0.12em}{0ex}}$ do it assigning values to what variable, $x\phantom{\rule{0.12em}{0ex}}$ or y?
don't really understand your explanation. thanks! $>$ can you provide an example?.
I did. It was ${y\phantom{\rule{0.12em}{0ex}}}^{4}=x\phantom{\rule{0.12em}{0ex}}$.

And, yes, only $b\phantom{\rule{0.12em}{0ex}}\right)$ is a function and what you describe with "vertical test" is exactly what I meant when I wrote that the relation has to be unique to be called a function. At every $x\phantom{\rule{0.12em}{0ex}}$ position there must be no more than just one assigned $y\phantom{\rule{0.12em}{0ex}}$ value.

To "see" if the relation is unique without plotting the relation, you would do the plotting in your mind. If you find just one x-value with two or more assigned y-values, than its not a function.

For example $|y\phantom{\rule{0.12em}{0ex}}|={x\phantom{\rule{0.12em}{0ex}}}^{2}$. You know that $e\phantom{\rule{0.12em}{0ex}}.g\phantom{\rule{0.12em}{0ex}}$. both $|+4|$ and $|-4|$ simplify to $+4$. This means that for $x\phantom{\rule{0.12em}{0ex}}=2\left(\to {x\phantom{\rule{0.12em}{0ex}}}^{2}=4\right)$ there are two possible values for $y\phantom{\rule{0.12em}{0ex}}\left(\to +4$ and $-4\right)$ and this contradicts the characterizing property of a function.  RASTA 20:22 Uhr, 09.01.2022

I got clearly!

I gave a value to $x\phantom{\rule{0.12em}{0ex}},$
- if I get 2 values of $y\phantom{\rule{0.12em}{0ex}}$ for the same $x\phantom{\rule{0.12em}{0ex}}$ then it is not a function.

So for the equation $A\phantom{\rule{0.12em}{0ex}}>{x\phantom{\rule{0.12em}{0ex}}}^{2}={y\phantom{\rule{0.12em}{0ex}}}^{2}$.
For $x\phantom{\rule{0.12em}{0ex}}=1,$ you can get $y\phantom{\rule{0.12em}{0ex}}=-1$ or 1. So it is not a function.

Equation $B\phantom{\rule{0.12em}{0ex}}$ gave me only one value so it's a function.

Because there is one y-value for the entered x-value, and only one.

$x\phantom{\rule{0.12em}{0ex}}=2,$ then, $y\phantom{\rule{0.12em}{0ex}}=4$ when $x\phantom{\rule{0.12em}{0ex}}=2$
$y\phantom{\rule{0.12em}{0ex}}={x\phantom{\rule{0.12em}{0ex}}}^{2}$
$y\phantom{\rule{0.12em}{0ex}}=4$

Vielen Danke.