A level further maths question

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#1
f(x)=(ax^2+bx+c)/(x^2+dx+e) a is not 0
f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)
show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?
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1 month ago
#2
(Original post by Emily~3695)
f(x)=(ax^2+bx+c)/(x^2+dx+e) a is not 0
f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)
show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?
I guess you've got to at least differentiate and consider the stationary points at x=2 and 0.5?
You could try making the fraction proper, but id guess this may not be necessary?
Last edited by mqb2766; 1 month ago
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#3
(Original post by mqb2766)
I guess you've got to at least differentiate and consider the stationary points at x=2 and 0.5?
You could try making the fraction proper, but id guess this may not be necessary?
This is year 12 Further maths so I don’t know how to differentiate fractions like this yet, in class we have found the stationary points by using the discriminant but I’m not sure how to apply that to this question
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1 month ago
#4
you just need to simplify it bring the bottom of the fraction up and (making the powers negative so mulitplying by (x^-2 dx^-1 e^-1) basic laws of indices,
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1 month ago
#5
(Original post by Emily~3695)
This is year 12 Further maths so I don’t know how to differentiate fractions like this yet, in class we have found the stationary points by using the discriminant but I’m not sure how to apply that to this question
I suppose you can sub x=2 into the quadratic to find the max value.
The intersection of that horizontal line with the function would have a single solution.
Cross multiply the quadratic fraction and max value and "solve" the resulting quadratic - single solution or discriminant.
Last edited by mqb2766; 1 month ago
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1 month ago
#6
(Original post by Emily~3695)
f(x)=(ax^2+bx+c)/(x^2+dx+e) a is not 0
f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)
show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?
not sure why you would attempt to find a discriminant equation when you mentioned f(x) has no (real?) roots..
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1 month ago
#7
Perhaps completing the square on both the numerator and denominator could help.

EDIT. Don't do this
Last edited by B_9710; 1 month ago
1
1 month ago
#8
(Original post by Emily~3695)
f(x)=(ax^2+bx+c)/(x^2+dx+e) a is not 0
f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)
show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?
I think people have been misinterpreting your question.

When you say "maximum at 2, minimum at 0.5", I presume you mean for all x, and that it achieves its bounds.

Your idea of using the discriminant for the maximum and minimum values will give you two equations.

Using the information for the y-intercept will give you a relationship between "c" and "e" - allowing you to eliminate "c".

And finally, since the function lies between 0.5 and 2, this means "a" can only take one possible value. Can you work out what that is?
Edit: See following.

Putting these bits of information together will allow you to derive the desired equation - "(b-0.5d)^2=(b-2d)^2=2e"

Note: No differentiaton is required.
Last edited by ghostwalker; 1 month ago
1
1 month ago
#9
(Original post by ghostwalker)
I think people have been misinterpreting your question.

When you say "maximum at 2, minimum at 0.5", I presume you mean for all x, and that it achieves its bounds.
I was wondering why this question didn't seem to make much sense as phrased
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1 month ago
#10
(Original post by davros)
I was wondering why this question didn't seem to make much sense as phrased
Can't recall seeing one like it on here before - doesn't mean there hasn't been one mind - and the "at 2" etc., does naturally lend itself to mean x=2, so somewhat misleading.
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#11
(Original post by ghostwalker)
Can't recall seeing one like it on here before - doesn't mean there hasn't been one mind - and the "at 2" etc., does naturally lend itself to mean x=2, so somewhat misleading.
Thank you for your help it makes more sense now the wording was definitely quite misleading which is why I think I was confused.
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1 month ago
#12
(Original post by Emily~3695)
Thank you for your help it makes more sense now the wording was definitely quite misleading which is why I think I was confused.
Oh <expletive deleted>!

I realised this morning that my solution is flawed.

The desired equation can be obtained if we have a=1, but just because the function's range is [0.5,2] (viz. bounded), this does not imply that a=1.
Last edited by ghostwalker; 1 month ago
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1 month ago
#13
(Original post by Emily~3695)
Thank you for your help it makes more sense now the wording was definitely quite misleading which is why I think I was confused.
Is what you've written exactly what the problem says, or have you rephrased it?

Is it taken from a book? Can you upload an image?
1
1 month ago
#14
(Original post by Emily~3695)
Thank you for your help it makes more sense now the wording was definitely quite misleading which is why I think I was confused.
Further checking shows that the desired equation is not necessarily true.

If you have a=3/2, b=7/3, c=3/2, d=2/3, e=1 (for example), then this set of values meets all the given criteria, but: (b-0.5d)^2=4e and (b-2d)^2=e

As davros asked, an image of the original question would be useful, if not a necessity, now.
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1 month ago
#15
(Original post by ghostwalker)
Further checking shows that the desired equation is not necessarily true.

If you have a=3/2, b=7/3, c=3/2, d=2/3, e=1 (for example), then this set of values meets all the given criteria, but: (b-0.5d)^2=4e and (b-2d)^2=e

As davros asked, an image of the original question would be useful, if not a necessity, now.
Nice detective work (PRSOM).
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#16
(Original post by ghostwalker)
Further checking shows that the desired equation is not necessarily true.

If you have a=3/2, b=7/3, c=3/2, d=2/3, e=1 (for example), then this set of values meets all the given criteria, but: (b-0.5d)^2=4e and (b-2d)^2=e

As davros asked, an image of the original question would be useful, if not a necessity, now.

This is the original question it’s pretty much exactly what I put
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1 month ago
#17
(Original post by Emily~3695)

This is the original question it’s pretty much exactly what I put
Your missed one vital fact from your post "a horizontal asymptote at y=1". This implies a=1 and the result follows.

Perhaps post an image of the question, next time.
Last edited by ghostwalker; 1 month ago
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#18
(Original post by ghostwalker)
Your missed one vital fact from your post "a horizontal asymptote at y=1". This implies a=1 and the rest follows.

Perhaps post an image of the question, next time.
Ah ok I didn’t even realise that when I was answering the question, thanks for your help
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1 month ago
#19
(Original post by Emily~3695)
Ah ok I didn’t even realise that when I was answering the question, thanks for your help
Without it you cannot deduce that a=1. I trust you understand why it means a=1: Divide numerator and denominator by x^2, and consider the limit as x goes to infinity.
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#20
(Original post by ghostwalker)
Without it you cannot deduce that a=1. I trust you understand why it means a=1: Divide numerator and denominator by x^2, and consider the limit as x goes to infinity.
Yes I do understand how to work out asymptotes I just wasn’t sure how to apply everything to this question. Thanks 😊
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