Trouble with question on rational functions.

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jc768
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Any help greatly appreciated here!

I am trying to solve the following problem:

Find a condition o c so that the function f(x)=(x+c)/(x^2-3x-c) has the whole real numbers as its range.


So far my thought process has been to find the range of c for which the function does not have any turning point, as if the function doesn't have any turning points, its range cant be restricted.

So I got the following:

https://ibb.co/j4NSsG1

If a turning point exists then when setting the function equal to a constant k it would have 1 solution.. but I'm not sure what to do next.
Last edited by jc768; 6 months ago
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mqb2766
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(Original post by jc768)
Any help greatly appreciated here!

I am trying to solve the following problem:

Find a condition o c so that the function f(x)=(x+c)/(x^2-3x+c) has the whole real numbers as its range.


So far my thought process has been to find the range of c for which the function does not have any turning point, as if the function doesn't have any turning points, its range cant be restricted.

So I got the following:

https://ibb.co/j4NSsG1

If a turning point exists then when setting the function equal to a constant k it would have 1 solution.. but I'm not sure what to do next.
Have you played a bit with the function?
https://www.desmos.com/calculator/hku3fawroc
As a hint, Id think about whether the (denominator) quadratic has any zeros, does completing the square help, ...

Note - unsure about the sign of c on the denominator?
Last edited by mqb2766; 6 months ago
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jc768
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(Original post by mqb2766)
Have you played a bit with the function?
https://www.desmos.com/calculator/hku3fawroc
As a hint, Id think about whether the (denominator) quadratic has any zeros, does completing the square help, ...

Note - unsure about the sign of c on the denominator?
Just realsied I have typed in the wrong function, should be

f(x)=(x+c)/(x^2-3x-c)
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mqb2766
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(Original post by jc768)
Just realsied I have typed in the wrong function, should be

f(x)=(x+c)/(x^2-3x-c)
Any thoughts (with the correction)?
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jc768
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(Original post by mqb2766)
Have you played a bit with the function?
https://www.desmos.com/calculator/hku3fawroc
As a hint, Id think about whether the (denominator) quadratic has any zeros, does completing the square help, ...

Note - unsure about the sign of c on the denominator?
The denominator has solutions when c>=-9/4 so this is when there are vertical asmptotes.
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mqb2766
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(Original post by jc768)
The denominator has solutions when c>=-9/4 so this is when there are vertical asmptotes.
Is that what you want? Solutions are zeros/roots on the denominator.
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jc768
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(Original post by mqb2766)
Is that what you want? Solutions are zeros/roots on the denominator.
Yes and they occur when the discriminant is >=0
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jc768
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(Original post by jc768)
Yes and they occur when the discriminant is >=0
Or by completing the square they are 1.5+-root(2.25+c)
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mqb2766
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(Original post by jc768)
Yes and they occur when the discriminant is >=0
Sounds ok, just put the explanation around it.
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mqb2766
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(Original post by jc768)
Or by completing the square they are 1.5+-root(2.25+c)
Yup, completing the square is often easier (hence suggestion).
You really dont need to know where they are, just that they exist.
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jc768
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(Original post by jc768)
Or by completing the square they are 1.5+-root(2.25+c)
Is it the case that a vertical asymptote proves a range exists for all real numbers?
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IQuitTSR
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(Original post by jc768)
So far my thought process has been to find the range of c for which the function does not have any turning point, as if the function doesn't have any turning points, its range cant be restricted.
Not true... The function e^x has no turning points, yet its range is not all of the real numbers

Firstly you need to check which conditions ensure there are horizontal or vertical asymptotes.
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jc768
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(Original post by TheTroll73)
Not true... The function e^x has no turning points, yet its range is not all of the real numbers

Firstly you need to check which conditions ensure there are horizontal or vertical asymptotes.
There is a horizontal asymptote at y=0 and vertical asymptotes occur when c>=-2.25
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mqb2766
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(Original post by jc768)
Is it the case that a vertical asymptote proves a range exists for all real numbers?
Not necessrily, but in this case you need a bit explanation.
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(Original post by jc768)
There is a horizontal asymptote at y=0 and vertical asymptotes occur when c>=-2.25
okay

you also know that the function is continuous in its domain (no breaks in the function except at asymptotes) and due to the limits as x approaches infinity and negative infinity are 0 there definitely needs to be a vertical asymptotes for the range to be all of the reals.

You need to find the right and left-handed limits as x approaches the asymptotes (for 1/x for example the left-handed limit (as x approaches 0) is negative infinity while the right-handed limit is infinity).

Now you should find any turning points and their nature. Again use the fact that f(x) is continuous.

If it is still unclear try sketching the graph for certain values of c to get an idea of how the function looks like.
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jc768
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(Original post by TheTroll73)
okay

you also know that the function is continuous in its domain (no breaks in the function except at asymptotes) and due to the limits as x approaches infinity and negative infinity are 0 there definitely needs to be a vertical asymptotes for the range to be all of the reals.

You need to find the right and left-handed limits as x approaches the asymptotes (for 1/x for example the left-handed limit (as x approaches 0) is negative infinity while the right-handed limit is infinity).

Now you should find any turning points and their nature. Again use the fact that f(x) is continuous.

If it is still unclear try sketching the graph for certain values of c to get an idea of how the function looks like.
I tried to locate the turning points in the original link I posted, but couldn't see where to go next.
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mqb2766
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(Original post by jc768)
I tried to locate the turning points in the original link I posted, but couldn't see where to go next.
You're ok now?
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jc768
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(Original post by mqb2766)
You're ok now?
No not at all, I could do with a worked solution to be honest.
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IQuitTSR
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(Original post by jc768)
I tried to locate the turning points in the original link I posted, but couldn't see where to go next.
Did you find the y-coordinate of the turning point? Is it a maxima, minima, or point of inflection? Does this section of the graph between the asymptotes have roots? What are limits as x approaches the coordinates vertical asymptotes to deduce if the range is all of the real numbers in those cases?

Also, are there values of c where you can show there is no turning point? You can use the limits as x approaches the coordinates vertical asymptotes to deduce if the range is all of the real numbers in those cases.
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jc768
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(Original post by TheTroll73)
okay

you also know that the function is continuous in its domain (no breaks in the function except at asymptotes) and due to the limits as x approaches infinity and negative infinity are 0 there definitely needs to be a vertical asymptotes for the range to be all of the reals.

You need to find the right and left-handed limits as x approaches the asymptotes (for 1/x for example the left-handed limit (as x approaches 0) is negative infinity while the right-handed limit is infinity).

Now you should find any turning points and their nature. Again use the fact that f(x) is continuous.

If it is still unclear try sketching the graph for certain values of c to get an idea of how the function looks like.
But what happens as x approaches the asymptotes depends o the value of c
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