anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
noenof the above
anonymous
  • anonymous
how?

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anonymous
  • anonymous
x+10 cancel out
anonymous
  • anonymous
got it?
anonymous
  • anonymous
I have never came across this type of function
anonymous
  • 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
  • inkyvoyd
It's the 4th,
inkyvoyd
  • inkyvoyd
Fourth selection.
anonymous
  • anonymous
no
inkyvoyd
  • inkyvoyd
Why not?
anonymous
  • anonymous
yeah it is
anonymous
  • anonymous
no, its 5th
inkyvoyd
  • inkyvoyd
X can be all real numbers but -10, cause that's undefined.
anonymous
  • anonymous
domain is all R
inkyvoyd
  • inkyvoyd
X can be 1, because when X is one it's defined as 2.
anonymous
  • anonymous
x+10 cancel out
inkyvoyd
  • inkyvoyd
Yes, so x=/=-10. X can be 1, because it's defined as 2.
inkyvoyd
  • inkyvoyd
Thus the domain is all real numbers except for 10.
anonymous
  • anonymous
I still do not understand
inkyvoyd
  • inkyvoyd
*-10
anonymous
  • 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
  • inkyvoyd
myko, I have no idea what you are saying, but I am 100% sure I'm correct.
anonymous
  • anonymous
if i repeat this n times, function will be undifind everywhere
anonymous
  • anonymous
kk, be my guest
inkyvoyd
  • inkyvoyd
x is undefined only at -10 in this peicewise function.
inkyvoyd
  • inkyvoyd
Thus it is defined for all points in the domain of real numbers *except* for -10
inkyvoyd
  • inkyvoyd
That is exactly what the 4th selection says.
anonymous
  • anonymous
yes, you wouls be right if no x+10 in nominator
anonymous
  • anonymous
what is a peicewise function
inkyvoyd
  • inkyvoyd
0/0, and a/0 for all real numbers a are both indeterminate forms. These forms are not defined.
anonymous
  • 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
  • anonymous
so now you have more problematic points, no?
inkyvoyd
  • inkyvoyd
myko, if x=/= to 2, then yes.
anonymous
  • anonymous
lol
inkyvoyd
  • inkyvoyd
and, chrissy, this is a piecewise function. http://en.wikipedia.org/wiki/Piecewise
anonymous
  • anonymous
i could add any (x-n) this way. Where n is Real nummber. So function f(x) would be undefind everywhere.....
anonymous
  • anonymous
piecewise function has nothing to do with (x+10) case
inkyvoyd
  • 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
  • anonymous
you are wrong
inkyvoyd
  • 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
  • anonymous
if i multiply and devide a function by the same thing, function stays unchanged!!!!!!!!!!!
inkyvoyd
  • inkyvoyd
You are multiplying by and dividing by x-n. You must add each time you do so, x=/=n.
inkyvoyd
  • inkyvoyd
Your logic is flawed because x=/=n.
anonymous
  • anonymous
(x+10)/[(x+10)(x-1)]= [(x+10)(x-2)]/[(x-2)(x+10)(x-1) ]
anonymous
  • anonymous
thats what i did
inkyvoyd
  • inkyvoyd
They are only the same when x=/=2
anonymous
  • anonymous
(x-2)=(x-2) for all x
inkyvoyd
  • 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
  • anonymous
even for 2
inkyvoyd
  • inkyvoyd
(x-2)/(x-2)=/=1 when x=2
anonymous
  • 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
  • inkyvoyd
Thus you can't multiply by (x-2)/(x-2) when x=2 because that's multiplying by 0/0
inkyvoyd
  • inkyvoyd
Multiplying both sides by (x-n) is invalid because each time you do it you add another condition.
inkyvoyd
  • inkyvoyd
Stop trolling.
anonymous
  • anonymous
anyway, writhe answer is 5th
anonymous
  • anonymous
see ya
inkyvoyd
  • 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
  • Muhammad_Nauman_Umair
ITS THE FOURH ONE
phi
  • 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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