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- anonymous

See screenshot attached

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- anonymous

##### 1 Attachment

- anonymous

noenof the above

- anonymous

how?

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## More answers

- anonymous

x+10 cancel out

- anonymous

got it?

- anonymous

I have never came across this type of function

- anonymous

for the function to be defind, the down part cant be 0. So we have to problematic points: 1 and -10. The point 1 was redefind. So -10 left. But if you look carefully, you see that you can simplifie you function a bit bu canceling terms that actualy dont afect the function, so -10, after all is not a bad point.:)

- inkyvoyd

It's the 4th,

- inkyvoyd

Fourth selection.

- anonymous

no

- inkyvoyd

Why not?

- anonymous

yeah it is

- anonymous

no, its 5th

- inkyvoyd

X can be all real numbers but -10, cause that's undefined.

- anonymous

domain is all R

- inkyvoyd

X can be 1, because when X is one it's defined as 2.

- anonymous

x+10 cancel out

- inkyvoyd

Yes, so x=/=-10. X can be 1, because it's defined as 2.

- inkyvoyd

Thus the domain is all real numbers except for 10.

- anonymous

I still do not understand

- inkyvoyd

*-10

- anonymous

the way you say it, i could make like this:
f(x) =( x-2) f(x)/(x-2)
so now also x = 2 is cut of the domain?

- inkyvoyd

myko, I have no idea what you are saying, but I am 100% sure I'm correct.

- anonymous

if i repeat this n times, function will be undifind everywhere

- anonymous

kk, be my guest

- inkyvoyd

x is undefined only at -10 in this peicewise function.

- inkyvoyd

Thus it is defined for all points in the domain of real numbers *except* for -10

- inkyvoyd

That is exactly what the 4th selection says.

- anonymous

yes, you wouls be right if no x+10 in nominator

- anonymous

what is a peicewise function

- inkyvoyd

0/0, and a/0 for all real numbers a are both indeterminate forms. These forms are not defined.

- anonymous

i guees you will agree that:
(x+10)/[(x+10)(x-1)]= [(x+10)(x-2)]/[(x-2)(x+10)(x-1) ]
no?

- anonymous

so now you have more problematic points, no?

- inkyvoyd

myko, if x=/= to 2, then yes.

- anonymous

lol

- inkyvoyd

and, chrissy, this is a piecewise function.
http://en.wikipedia.org/wiki/Piecewise

- anonymous

i could add any (x-n) this way. Where n is Real nummber. So function f(x) would be undefind everywhere.....

- anonymous

piecewise function has nothing to do with (x+10) case

- inkyvoyd

No, because x would have to be unequal to n. If you added an infinite amount of x=/=n and x-n the result would be contradictory.

- anonymous

you are wrong

- inkyvoyd

No, I am not. You can't just add (x-n) to the numerator and denominator and infinite amount of times, because x=/=n.

- anonymous

if i multiply and devide a function by the same thing, function stays unchanged!!!!!!!!!!!

- inkyvoyd

You are multiplying by and dividing by x-n. You must add each time you do so, x=/=n.

- inkyvoyd

Your logic is flawed because x=/=n.

- anonymous

(x+10)/[(x+10)(x-1)]= [(x+10)(x-2)]/[(x-2)(x+10)(x-1) ]

- anonymous

thats what i did

- inkyvoyd

They are only the same when x=/=2

- anonymous

(x-2)=(x-2) for all x

- inkyvoyd

if x=2, you are multiply by 0/0, which is an indeterminant form. You can not multiply by an indeterminant form and expect a consistent answer.

- anonymous

even for 2

- inkyvoyd

(x-2)/(x-2)=/=1 when x=2

- anonymous

thinking this way:
i could add any (x-n) this way. Where n is Real nummber. So function f(x) would be undefind everywhere.....

- inkyvoyd

Thus you can't multiply by (x-2)/(x-2) when x=2 because that's multiplying by 0/0

- inkyvoyd

Multiplying both sides by (x-n) is invalid because each time you do it you add another condition.

- inkyvoyd

Stop trolling.

- anonymous

anyway, writhe answer is 5th

- anonymous

see ya

- inkyvoyd

Dude, stop trolling. Your logic is flawed, you now it too, and you continue to troll. People like you aren't welcome here.

- Muhammad_Nauman_Umair

ITS THE FOURH ONE

- phi

Just to be clear. There are 2 things going on here.
First, we have a piece-wise definition of a function
That means we have different definition depending on the value of the independent variable.
In this case the function is defined to be 2 when x=1.
When x≠1, the function is defined by the expression
\[ \frac{x+10}{(x+10)(x-1)} \]
If x= -10, the denominator will be 0, the numerator will be 0, and we have 0/0 (zero divided by zero) which is undefined. So x= -10 must be excluded from the domain to prevent this nasty result.
Notice x=1 would result in 11/0 which is also undefined. This would be a problem except the definition of the function does not use this expression when x=1. It defines the function to be 2 when x=1. So x=1 is ok.
Now it is true we could cancel (x+10) from the definition:
\[ \frac{\cancel{(x+10)}}{\cancel{(x+10)}(x-1)} = \frac{1}{x-1}, \text{x≠ -10}\]
but the two expressions are not truly identical unless we take note that x≠ -10
It is a subtle point that is often ignored.
So, if your function definition is a ratio of expressions, anything that makes the denominator zero is not in the domain (UNLESS that x value is added back in using a piece-wise definition)

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