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How can I plug in a value for $x$ in $y = f(x)$ and get a result for $y$?

What does the $f$ do in the equation?

I know it stands for a function, but does it actually represent a value that I should plug in?

2 Answers 2

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$f$ does indeed denote a function, and expressing it as $f(x)$ makes it clear that $f$ is a function of x.
$f$, or $y = f(x)$, does not represent a value you "plug in": rather, to evaluate the value of $y = f(x)$ at a given value of $x$, you need to "plug" that given value of $x$, say $a$, into $f(x)$; in other words, you need to evaluate $f(a)$.

More details:

$y = f(x)$ means that $y$ is a function of x. To evaluate the function for a given x, in order to determine the value of $y = f(x)$, you need to know the function:

e.g. Suppose $y = f(x)$ where $f(x) = x^2 +7$. Then, if you want to "plug in" a value for $\bf{x}$, (also known as "evaluating y or f(x) at a given value $x$), say "x = 3": that means for $x = 3$, we have $y = f(3) = 3^2 + 7 = 9 + 7 = 16$. f(x) and "y" are sometimes used interchangeably, but f(x) is more explicit about being a function of $x$.

"$x$" is the value at which you evaluate $y = f(x)$.

"$y$" is often used instead of f(x), when it's clear that $y$ is a function of $x$, especially, for example, when the y-coordinate of a set of points in the Cartesian plane is expressed as a function of the x-coordinates of the set of points.

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    oh, that makes so much more sense, thanks!2011-05-31
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    @Studiohack: your welcome! Does that answer your question?2011-05-31
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    it does @amWhy, thanks again! :D2011-05-31
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    @Chandru: I'm aware, I'm a mod elsewhere ;)2011-05-31
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    @Studiohack: Sorry, i saw your profile and it seems that you have joined today. So i asked you to do this. PLease don't mind.2011-05-31
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    @Chandru I don't mind, I understand. I was seen 11 minutes ago, and joined 9 months ago.2011-05-31
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    @studiohack: OOPS, I don't scrutinize the details properly :)2011-05-31
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Yes it does represent a value.

  • Example if you take your $f(x)=x$ then when you plug in $1$, the output $y=f(1)=1$.

  • If you take your $f(x)=x^{2}$, then when you plug $x=2$, your output is $2^{2}=4$.

  • If you take $f(x)=|x|$, then when you put $x=-2$, your output is $|-2|=2$.

For more information, i suggest you to read this Wikipedia page:

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    about the broken link, see my comment in http://math.stackexchange.com/questions/42257/minimal-polynomial2011-05-31
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    @lhf: Thanks. I have done.2011-05-31
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    thanks for the answer @Chandru, noticed you expanded your answer, FGIW2011-05-31
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    @studiohack: Yes, I did. Welcome :)2011-05-31