# Trouble with question on rational functions.

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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.

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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#2

(Original post by

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.

**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.

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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(Original post by

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?

**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?

f(x)=(x+c)/(x^2-3x-c)

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#4

(Original post by

Just realsied I have typed in the wrong function, should be

f(x)=(x+c)/(x^2-3x-c)

**jc768**)Just realsied I have typed in the wrong function, should be

f(x)=(x+c)/(x^2-3x-c)

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**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?

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#6

(Original post by

The denominator has solutions when c>=-9/4 so this is when there are vertical asmptotes.

**jc768**)The denominator has solutions when c>=-9/4 so this is when there are vertical asmptotes.

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(Original post by

Is that what you want? Solutions are zeros/roots on the denominator.

**mqb2766**)Is that what you want? Solutions are zeros/roots on the denominator.

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(Original post by

Yes and they occur when the discriminant is >=0

**jc768**)Yes and they occur when the discriminant is >=0

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#9

(Original post by

Yes and they occur when the discriminant is >=0

**jc768**)Yes and they occur when the discriminant is >=0

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#10

(Original post by

Or by completing the square they are 1.5+-root(2.25+c)

**jc768**)Or by completing the square they are 1.5+-root(2.25+c)

You really dont need to know where they are, just that they exist.

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(Original post by

Or by completing the square they are 1.5+-root(2.25+c)

**jc768**)Or by completing the square they are 1.5+-root(2.25+c)

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#12

(Original post by

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.

**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.

Firstly you need to check which conditions ensure there are horizontal or vertical asymptotes.

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(Original post by

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.

**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.

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#14

(Original post by

Is it the case that a vertical asymptote proves a range exists for all real numbers?

**jc768**)Is it the case that a vertical asymptote proves a range exists for all real numbers?

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#15

(Original post by

There is a horizontal asymptote at y=0 and vertical asymptotes occur when c>=-2.25

**jc768**)There is a horizontal asymptote at y=0 and vertical asymptotes occur when c>=-2.25

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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(Original post by

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.

**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.

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#17

(Original post by

I tried to locate the turning points in the original link I posted, but couldn't see where to go next.

**jc768**)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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(Original post by

You're ok now?

**mqb2766**)You're ok now?

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#19

**jc768**)

I tried to locate the turning points in the original link I posted, but couldn't see where to go next.

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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**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.

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