moongazer
how do you get the domain of this:
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moongazer
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how do you find it? step-by-step please :)
moongazer
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√(x^2 - 9x +20)/√(x^2 - 5x + 6)
I already know the domain. But, how do you get it?
ganeshie8
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first workout contraints for squareroot : the thing inside squareroot cannot be negative
ganeshie8
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next work out constraints for denominator : denominator cannot be equal to 0
ganeshie8
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consider numerator first : \(\sqrt{x^2 - 9x +20}\)
ganeshie8
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can u work out domain for that ?
moongazer
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{x|x>=5} is that right?
ganeshie8
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somewhat right :)
lets see how to solve
ganeshie8
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x^2-9x+20 >= 0
x^2-4x-5x+20 >=0
x(x-4) -5(x-4) >= 0
(x-4)(x-5) >= 0
x <= 4 or x >= 5
ganeshie8
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thats one constraint for numerator, lets find out the constraints for denominator
moongazer
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ohhhh
moongazer
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I think I am starting to understand it :)
ganeshie8
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good :)
since, denominator is also under radical, it must be >=0 :
x^2 - 5x + 6 >= 0
x^2-3x-2x + 6 >= 0
x(x-3) -2(x-3) >= 0
(x-2)(x-3) >= 0
x <= 2 or x >= 3
ganeshie8
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with those two things, we finished with the first step of finding constraints for radicals
ganeshie8
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so far we have this :
x <= 4 or x >= 5
x <= 2 or x >= 3
moongazer
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yup, I understand it :)
ganeshie8
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next one is denominator can never equal to 0. so,
x^2 - 5x + 6 \(\ne\) 0
x^2-3x-2x + 6 \(\ne\) 0
x(x-3) -2(x-3) \(\ne\) 0
(x-2)(x-3) \(\ne\) 0
x \(\ne\) 2 or x \(\ne\) 3
moongazer
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w8 Isn't it that it should be:
x >= 4 or x >= 5
x >= 2 or x >= 3
???
ganeshie8
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lets combine all constraints and make a meaningful constraint
ganeshie8
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which one ?
moongazer
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(x-4)(x-5) >= 0
(x-2)(x-3) >= 0
ganeshie8
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oh i got ur question, il give quick explanation
ganeshie8
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you comfortable with parabola graph ?
moongazer
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I think so
ganeshie8
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il show u in graph why its x <=4 , x >=5
moongazer
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I know few things about parabola
moongazer
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ok
ganeshie8
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im sketching this parabola :
(x-4)(x-5) >= 0
ganeshie8
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it intersect x axis at 4, 5 right ?
ganeshie8
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|dw:1348496600444:dw|
ganeshie8
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between 4 and 5, it is sinking down into x axis eh ?
ganeshie8
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its becoming NEGATIVE
ganeshie8
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|dw:1348496705657:dw|
moongazer
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why did it intersect in x axis at 4, 5?
ganeshie8
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when u factored you got : (x-4)(x-5)
ganeshie8
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that means it intersects x-axis at 4 and 5
moongazer
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ohh, ok
what if it says (x-4)(x+5)
does it intersect at 4 and -5 ?
ganeshie8
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thats right,
when it says, (x-4)(x+5) = 0
then it intersects at 4 and -5
ganeshie8
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cuz for it to equal to 0, one of both of the factors must equal to 0
moongazer
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|dw:1348498226449:dw|
moongazer
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so the graph is like that?
ganeshie8
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perfect, thats (x-4)(x+5)
moongazer
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and the domain is x>=4,x<=-5
is it correct?
ganeshie8
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thats right ! you became expert ;p
moongazer
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for (x-4)(x+5)
moongazer
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Thanks! I know understand it. :)
let's now continue with our previous discussion :)
ganeshie8
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great ! so far what we have
moongazer
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next one is denominator can never equal to 0. so,
x^2 - 5x + 6 ≠ 0
x^2-3x-2x + 6 ≠ 0
x(x-3) -2(x-3) ≠ 0
(x-2)(x-3) ≠ 0
x ≠ 2 or x ≠ 3
ganeshie8
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yeah we almost done
ganeshie8
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lets put down all the 3 constraints we got in one place :
ganeshie8
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1) numerator radical : x <= 4 or x >= 5
2) denominator radical : x <= 2 or x >= 3
3) denominator cant be 0 : x ≠ 2 or x ≠ 3
moongazer
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I understand it. What's next?
ganeshie8
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see 2) and 3)
ganeshie8
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can we combine them as one, like this :
2) 3) : x < 2 or x > 3
moongazer
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I'll be back just continue.
moongazer
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someone is calling me
ganeshie8
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ok after combining 2) and 3) we are left with :
1) numerator radical : x <= 4 or x >= 5
2) x < 2 or x > 3
this is the last step. see if u can figure out
ganeshie8
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out of x < 2 & x <= 4,
x < 2 is more restrictive, so we have this in our domain : \(\color{green}{x < 2}\)
ganeshie8
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out of x > 3 & x <= 4
x > 3 , x <= 4 , which is same as : \(\color{green}{3<x\le 4}\)
ganeshie8
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and of course the last one, \(\color{green}{x >= 5}\)
moongazer
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sorry for being away, I'm here again :)
ganeshie8
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np :) see if above stuff makes sense
moongazer
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it makes sense
but why in the final answer they used "or" not "and" "," "then" or anything else
dom K: {x|x<2 or 3<x<=4 or x>=5}
ganeshie8
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yeah we always combine domain with "or"
why ? cuz its meaningful.
let me ask u a q
ganeshie8
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suppose we write : x < 2 and x >=5
moongazer
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for the domain it is ALWAYS "or" ?
ganeshie8
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is that possible ? can x be BOTH less than 2 and greater than 5 at the SAME time ha ?
ganeshie8
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yes always "or"
moongazer
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yes, it is not possible
ganeshie8
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is "1" less than 2 and greater than "5" ?
ganeshie8
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yeah, but if we put it like this :
either "1" is less than 2 or greater than "5"
ganeshie8
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how meaningful it is, so while phrasing domain also, we use commonsense aswell :)
moongazer
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yes, Thank you very much for helping me :)
ganeshie8
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np i learned as well. i forgot these long back in college... i enjoyed going thru these again.. thnks you too :)