Question

Finding the excluded values ... MEDAL!!!

Answer 1

\[\large \frac{ 7r + 3 }{ 2(7r +6) }\]

Answer 2

Firstly, do you know what an excluded value is?

Answer 3

Yes when the denominator is equal to 0

Answer 4

Exactly, so we set \[2(7r+6)=0\]and solve for r. At this point, the fraction will be undefined so we exclude this value of r from our domain. In this case there is only one value, there may be more than one in other cases.

Answer 5

The answer key says that the excluded values are:
\(-\frac{6}{7}\) and \(6\)

Answer 6

I got \(-\frac{6}{7}\) but I did not get 6.

Answer 7

was this a function you found after composing a function with another function?

Answer 8

Unless I've completely lost it, I disagree with that answer. Are you sure this question is correct?

Answer 9

Should've said, it's the 6 I disagree with, not the -6/7. How do you do inline LaTeX btw? I can't for the life of me work it out.

Answer 10

yes? can you tell us the functions you had
and what composition you were to find

Answer 11

for example if you were to find f(g(x))
and g(a) isn't defined then a wouldn't be in the domain of f(g(x))

Answer 12

We had to simplify the fraction
\[\large \frac{ 7r^2 - 39r - 18 }{ 14r^2 - 72r - 72 }\]

Answer 13

okay factor the bottom and set equal to 0 and solve

Answer 14

you find when the bottom is 0 before simplifying

Answer 15

have to go
tom is here

Answer 16

I did. And so I had \(\large 2(7r + 6)\) as the denominator.

Answer 17

Bye @freckles, thanks for your help - wouldn't have guessed it had been simplified! What @freckles said was to factorise the denominator and solve for f(r)=0 \[14r^2-72r-72=0\]What values of r satisfy this?

Answer 18

Do you mean to use the quadratic equation?

Answer 19

Yep, that's one way and probably the easiest in this case.

Answer 20

But didn't I factorize that already?

Answer 21

I think you have, but you cancelled it down which meant you lost the 6.

Answer 22

Oh yes that's right

Answer 23

I had:
\[\large \frac{ (r-6)(7r+3) }{ (r-6)(7r +6) }\]

Answer 24

So you are saying that before I cancel out the terms that need to be canceled out, I should get get the excluded value ?

Answer 25

Okay I got the answer to my question from youtube. Thanks for your help :)

Answer 26

remember, you can divide something by itself and get 1 *only if not zero*
in other words, when r=6, you have 0/0 and that is not 1. so you have to exclude r=6

Answer 27

Yeah that looks more like it. When finding solutions, you have to be so careful you aren't potentially dividing by zero or things could break or you could lose roots. A great example is the good old 2=1 proof:
\[a=b\]\[a^2=ab\]\[a^2-b^2=ab-b^2\]\[(a+b)(a-b)=b(a-b)\]\[* a+b=b\]\[2a=a\]\[2=1\]See the line with the asterisk, well to get to that line I divided by a-b but because a=b I've just divided by 0 (a-a).

You shouldn't be cancelling anything here, leave it as is. If you do divide by something, you have to take it into consideration. This question isn't a good example to explain that with, but this is really important with solving some equations, like this:

http://tiny.cc/latexquestion

Factor and consider each case, don't divide blindly.