Brudor2000
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Is there a specific method to find domain for a function.
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Brudor2000
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I know for some functions drawing the graph helps but what about complicated functions for which drawing a graph would be a pain in exams?
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RDKGames
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(Original post by Brudor2000)
I know for some functions drawing the graph helps but what about complicated functions for which drawing a graph would be a pain in exams?
Give an example of a 'complicated' function.
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Brudor2000
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(Original post by RDKGames)
Give an example of a 'complicated' function.
well i cant't think of or find one rn. even if function is not complicated is there another way to find the domain other than drawing a graph.
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Kota Dagnino
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The main thing to do is see if there are any values where f(x) doesn't exist. For example, if you have something like
1/(x-2)

we know that the denominator of a fraction can't be equal to zero. This means that x-2 must not be equal to 0.
If we actually think about what happens at x=2, we understand that the function approaches infinity, and therefore has an asymptote there.
Look at the attached file to see the asymptote.
Attached files
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Brudor2000
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(Original post by Kota Dagnino)
The main thing to do is see if there are any values where f(x) doesn't exist. For example, if you have something like
1/(x-2)

we know that the denominator of a fraction can't be equal to zero. This means that x-2 must not be equal to 0.
If we actually think about what happens at x=2, we understand that the function approaches infinity, and therefore has an asymptote there.
Look at the attached file to see the asymptote.
therefore the domain of function 1/(x-2) is..?
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Anonymouspsych
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(Original post by Brudor2000)
well i cant't think of or find one rn. even if function is not complicated is there another way to find the domain other than drawing a graph.
well you don't always have to draw a graph if you know the general shape of the function. For example if you have something like f(x)= x-5/x+1, then the domain would be all real values of x apart from x=-1 (as then you would get -6/0 which is undefined).

But Generally drawing a graph will help
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Kota Dagnino
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so the domain is (-infinity,2) and (2, infinity)
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Brudor2000
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(Original post by Kota Dagnino)
so the domain is (-infinity,2) and (2, infinity)
could you give that in terms of f(x) >... or f(x)<...
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Brudor2000
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(Original post by Anonymouspsych)
well you don't always have to draw a graph if you know the general shape of the function. For example if you have something like f(x)= x-5/x+1, then the domain would be all real values of x apart from x=-1 (as then you would get -6/0 which is undefined).

But Generally drawing a graph will help
ohh ok. so what would the domain be for the eqn u have given?
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Kota Dagnino
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Here is another example,
Attached files
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RDKGames
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(Original post by Brudor2000)
well i cant't think of or find one rn. even if function is not complicated is there another way to find the domain other than drawing a graph.
Yes, just look at what the function is made up from to start with. Certain functions like \ln x are only valid for x&gt;0 so if your function is something like x^2\ln (x+1) then this is right-away valid for x&gt;-1 (or stricter).
Otherwise, you can get rational functions like \dfrac{1}{(x-1)(x-2)} which are undefined at only x=1 and x=2 so this function is valid for all reals except those two numbers.

The product of the two functions mentioned, \dfrac{x^2 \ln (x+1)}{(x-1)(x-2)} would constitute to a domain that is x&gt;-1 except x=1 and x=2.

So really, you should know by heart the domain of your basic special functions in C3 like \ln x, and that the denominators cannot ever be equal to 0, then apply these to more complicated functions which combine the two.


Forgot what's in C3 so I just used these two examples to illustrate the typical procedure.
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Brudor2000
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what about range of an equation when they have given a domain such as x is all real number???
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Kota Dagnino
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Basically, if you have a domain that looks like this (x,X) then its range will be (f(x), f(X)).
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RDKGames
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(Original post by Kota Dagnino)
Basically, if you have a domain that looks like this (x,X) then its range will be (f(x), f(X)).
That's not true.

Example: f(x)=x^2 for -2 \leq x \leq 1
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Brudor2000
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(Original post by Kota Dagnino)
Basically, if you have a domain that looks like this (x,X) then its range will be (f(x), f(X)).
yh but what if it was not definite like (x,X) and it was something like x E R ( x is all real number)
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Kota Dagnino
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Here is another example witten another way
Attached files
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Kota Dagnino
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if they say that the domain is all real numbers, then the range will also be all real numbers
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Kota Dagnino
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(Original post by RDKGames)
That's not true.

Example: f(x)=x^2 for -2 \leq x \leq 1
We are talking in the case of linear functions
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Brudor2000
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(Original post by Kota Dagnino)
if they say that the domain is all real numbers, then the range will also be all real numbers
wat about for exponential functions, trig functions etc??
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