Graphs of Rational Functions Watch

Y12_FurtherMaths
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#1
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Can someone help me with 8 please. I’ve got the graph drawn but I’m unsure where to go next. Thanks
https://imgur.com/a/XvUzV2R
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mqb2766
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It would make a lot more sense if it was
(x^2-3x+2)/(x^2+x-2)
As it stands, its not really well posed.

(Original post by Y12_FurtherMaths)
Can someone help me with 8 please. I’ve got the graph drawn but I’m unsure where to go next. Thanks
https://imgur.com/a/XvUzV2R
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Y12_FurtherMaths
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(Original post by mqb2766)
It would make a lot more sense if it was
(x^2-3x+2)/(x^2+x-2)
As it stands, its not really well posed.
Not sure what you mean. How does this help?
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mqb2766
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I suspect the question is wrong.
What is the book it comes from?
(Original post by Y12_FurtherMaths)
Not sure what you mean. How does this help?
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Y12_FurtherMaths
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(Original post by mqb2766)
I suspect the question is wrong.
What is the book it comes from?
A level further mathematics for AQA student book 1
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Y12_FurtherMaths
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(Original post by mqb2766)
I suspect the question is wrong.
What is the book it comes from?
This is the answer in the book
https://imgur.com/a/8Fj963d
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mqb2766
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Ok, I was wrong, it wants you to do a bit more analysis than I thought.
What values did it give for a and b
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Y12_FurtherMaths
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(Original post by mqb2766)
Ok, I was wrong, it wants you to do a bit more analysis than I thought.
What values did it give for a and b
a=b=2
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mqb2766
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f(3) = 2 and the function decreases for a time before that.
f(2) = 2 ...
so yes it works, a=b=2.

I was reading it like the interval 0<f(x)<a was the range or image and there isn't a solution for f(x)=1, say. Really should have thought of it as the codomain where not all values are invertible.

(Original post by Y12_FurtherMaths)
a=b=2
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Y12_FurtherMaths
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(Original post by mqb2766)
f(3) = 2 and the function decreases for a time before that.
f(2) = 2 ...
so yes it works, a=b=2.

I was reading it like the interval 0<f(x)<a was the range or image and there isn't a solution for f(x)=1, say. Really should have thought of it as the codomain where not all values are invertible.
Sorry but I’m not sure how you work out that a=b=2?
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mqb2766
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Before x=3 (given), f(x) is locally increasing so the upper bound on f(x) must be f(3) = 2 = a.
There is a minimum around x=2.? because f(1) = inf (asymptote, denominator=0) and f(x) -> 3 as x-> inf.
So there must be an x~2 such that f(x)=2 and this will be "b" as 0<f(x)<2 on the interval (b,3). Solve for x where f(x)=2 and its x=2=b.

(Original post by Y12_FurtherMaths)
Sorry but I’m not sure how you work out that a=b=2?
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