If = (x + x, for all x, what is the value of ... ...
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Posted: 05/14/2012 00:22
What do the > signs do, and where did all the two's come from?
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Posted: 05/14/2012 01:04
What do the > signs do, and where did all the two's come from?
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Posted: 10/05/2012 12:33
What happened to the x on the outside of the parentheses of the function ?
When I worked it out I got (x+2)(x+2) - (x+2)(x-2). Then I foiled...
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Posted: 04/16/2013 19:06
Ngan, I answered this before. Please follow the thread. Think of symbols in defined function questions as ƒ(..).
So ‹x› = ƒ(x) = (x+2)x
and:
‹x+2› = ƒ(x+2) = {(x+2) + 2} (x+2) = (x + 4) (x +2)
‹x-2› = ƒ(x - 2) = {(x-2) + 2} (x-2) = x (x-2)
I think you forgot to plug in (x+2) for the x inside the parentheses.
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Posted: 04/16/2013 19:06
Thad, think of symbols in defined function questions as ƒ(..)
So ‹x› = ƒ(x) = (x+2)x
and
‹x + 2› = ƒ(x+2) = {(x+2) + 2} (x+2) = (x + 4) (x +2)
‹x - 2› = ƒ(x - 2) = {(x-2) + 2} (x-2) = x (x-2)
etc.
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Posted: 10/04/2012 21:32
I understand where the initial ([x+2] + 2) comes from in the beginning of the solution, but where does the [x+2] come from after that? It seemed to me that you are substituting with (x+ 2)x. I had it down (for the first half) ([x+2] + 2) - ([x+2] - 2)
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Posted: 10/05/2012 14:34
Lola, you have to plug in (x+2) for every x you see. Please see the thread above yours.
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Posted: 04/02/2013 17:30
I agree with this comment and do not find the explanation clear. Could you break it down more?
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Posted: 04/16/2013 19:07
Tom, please see the thread. Think of symbols in defined function questions as ƒ(..)
So think of ‹x› = ƒ(x) = (x+2)x
and so,
‹x + 2› = ƒ(x+2) = {(x+2) + 2} (x+2) = (x + 4) (x +2)
‹x - 2› = ƒ(x - 2) = {(x-2) + 2} (x-2) = x (x-2)
Now it's just algebra, subtracting the second expression from the first one.
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If = (x + x, for all x, what is the value of ... ...
Posted: 07/26/2013 23:04
I can't find a reason why we substitute the ( 2nd X)with (x+2)
I saw the thread, I don't know why we don't use ((x+2)x))+2 - ((x+2)x) - 2
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Posted: 05/10/2015 14:14
Reply: exactly! Something isn't right here!
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Posted: 05/10/2015 14:57
Hi Tony,
The problem is psychologically confusing because the formula is written in terms of the variable x and the expressions being plugged into the formula are written in terms of the same variable, x.
Changing the variable in the formula to, say, z will make the substitution clearer:
= (z + 2)z
Now, to calculate , we replace every z in the formula = (z + 2)z with x + 2:
= ([x + 2] + 2)[x + 2] = (x +4)(x + 2) = x^2 + 6x + 8.
Similarly, to calculate , we replace every z in the formula = (z + 2)z with x - 2:
= ([x - 2] + 2)[x - 2] = (x)(x - 2) = x^2 - 2x.
Subtracting these two results gives the answer:
- =
[x^2 + 6x + 8] - [x^2 - 2x] =
x^2 + 6x + 8 - x^2 + 2x =
8x + 8 =
8(x + 1)
Nova Press
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Posted: 07/27/2013 13:19
Sudad, as has been explained in the thread, think of symbols in defined function questions, like <> here as ƒ(..).
So ‹x› = ƒ(x) = (x+2)x
and:
‹x+2› = ƒ(x+2) = {(x+2) + 2} (x+2) = (x + 4) (x +2)
‹x-2› = ƒ(x - 2) = {(x-2) + 2} (x-2) = x (x-2).
In functions of x, every time you see x, you need to replace it with the new term inside the parentheses.
