Question

See screenshot attached

Answer 1

noenof the above

Answer 2

how?

Answer 3

x+10 cancel out

Answer 4

got it?

Answer 5

I have never came across this type of function

Answer 6

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.:)

Answer 7

It's the 4th,

Answer 8

Fourth selection.

Answer 9

no

Answer 10

Why not?

Answer 11

yeah it is

Answer 12

no, its 5th

Answer 13

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

Answer 14

domain is all R

Answer 15

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

Answer 16

x+10 cancel out

Answer 17

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

Answer 18

Thus the domain is all real numbers except for 10.

Answer 19

I still do not understand

Answer 20

*-10

Answer 21

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?

Answer 22

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

Answer 23

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

Answer 24

kk, be my guest

Answer 25

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

Answer 26

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

Answer 27

That is exactly what the 4th selection says.

Answer 28

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

Answer 29

what is a peicewise function

Answer 30

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

Answer 31

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

Answer 32

so now you have more problematic points, no?

Answer 33

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

Answer 34

lol

Answer 35

and, chrissy, this is a piecewise function.

http://en.wikipedia.org/wiki/Piecewise

Answer 36

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

Answer 37

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

Answer 38

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.

Answer 39

you are wrong

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

Your logic is flawed because x=/=n.

Answer 44

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

Answer 45

thats what i did

Answer 46

They are only the same when x=/=2

Answer 47

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

Answer 48

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.

Answer 49

even for 2

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

Stop trolling.

Answer 55

anyway, writhe answer is 5th

Answer 56

see ya

Answer 57

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

Answer 58

ITS THE FOURH ONE

Answer 59

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)