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**Au101****Member**- Registered: 2010-12-01
- Posts: 348

Hi guys

I have a question about the first part of one of my linear interpolation questions.

Show that the largest possible root of the equation:

Lies in the interval [2,3].

I know how to prove that the root lies in this interval - one can simply let x = 2 and then let x = 3 and show that there is a change of sign. What I'm not sure about is quite how to show that this is the largest possible root. It seems fairly intuitively so from the graph, but I'm not sure about a good proof .

Thanks .

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,787

hi Au101,

You can get the roots easily and they are both to the left of the interval.

So the turning points are to the left of the interval.

You can state (without proof) that the general shape of a cubic with positive x cubed term is up, down, then up again.

Taken together this means the root in the interval must be the rightmost.

Bob

*Last edited by bob bundy (2011-02-25 10:18:37)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Au101****Member**- Registered: 2010-12-01
- Posts: 348

Ah that's perfect. Thank you once again Bob!

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**Au101****Member**- Registered: 2010-12-01
- Posts: 348

Hi

I have another, similar question:

Show that the equation

Has one positive and two negative roots.

Can I just sub in successive integers and show where the roots occur?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,787

hi Au101,

Well I suppose you could, but this might be considered a bit like finding by exhaustion; ie not so good if you don't find them fairly quickly.

In your last problem, it was something you said that pushed me towards the differentiation.

You said

It seems fairly intuitively so from the graph

and so I thought 'How about trying graph sketching techniques'. With A level, sketching is considered 'superior', because you are being more analytic I suppose, than if you just plot points. You are finding out important properties of the graph without resorting to calculating a load of points.

Plus: If you plot points someone could always argue that the graph might be doing odd things between your points; you might miss an asymptote where the graph zooms off to infinity for instance.

So I thought about finding the turning points and the quadratic was so nice to factorise that I felt I was on the right track.

And it happens again here.

So turning points when x = + or - 2 with y = (8 - 24 - 7.2) < 0 and ( -8 +24 - 7.2) > 0 respectively.

Then put x = 0 to get the crossing point with the y axis (0,-7.2) plus the usual 'shape of a cubic' argument and you've got enough to argue that the graph must cross the x axis first in the negatives, then have a maximum at x = -2, then cross the x axis again before going through (0,-7.2), then the minimum at x = + 2 before finally crossing the x axis for the last time on the positive x axis.

How are you with graph sketching generally? I can give you a list of things to look for if you want.

Bob

*Last edited by bob bundy (2011-02-26 10:46:47)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Au101****Member**- Registered: 2010-12-01
- Posts: 348

Oooh that's very good, thanks again Bob

In all honesty, graph sketching is probably one of my weakest areas, so that would be very nice, thanks .

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,787

hi Au101

Ok then: Graph Sketching by BB

Why?

There are often reasons why you want to know what a graph looks like. Your example here is a good one. Suppose you are trying to find any roots that are positive (you know from the problem that negatives won't apply). Knowing that there is only one positive root means you don't have to go searching for another.

And if you are trying to get an area by integration, you should always have a sketch first in case some of the graph dips below the axis. If you wanted the area between

and the x axis between x = 0 and x = 5

you would get the wrong answer if you just do one integral with these limits.

Part of the curve is below the x axis and will reduce the area result. You have to sketch the curve; find the negative bit; compute the three areas separately, making the middle bit positive; and then add them up.

And sketching means considering every bit of the curve whereas plotting means you only know what the co-ordinates are at specific values. If you wanted to know what

is like and you calculated y for x = 20, 40, 60, 80, 100,120 ...... then you would miss the strange behaviour at x = 90 altogether.

Even if the question required accurate plotting in order to read off some values (say, where two curves cross that cannot be done algebraically), it would still be a good idea to sketch first so that you know roughly where to plot the accurate version.

So what should you consider?

(i) Where does the curve cross the x axis? This would mean finding the roots.

(ii) Where does the curve cross the y axis? Easier, as you only have to put x = 0.

(iii) How does y behave when x tends to infinity? How does y behave when x tends to - infinity.

(iv) Does the curve have turning points? Are they maxima, minima or points of inflexion? What is the y value for each x; or, at least, is y positive or negative in these cases.

(v) What symmetry does the curve have? (in an axis for example or about the origin by rotation) Do you know about odd and even functions?

(vi) Does the curve have any asymptotes? ie Does the curve approach a particular line without touching it? Asymptotes may be parallel with an axis but may also be other straight lines. If y goes to infinity as x approaches a certain value, does the same thing happen as you approach this value from the left and from the right? (Think about the tan curve. As you go up to 90 the tan goes to + infinity, but if you come down towards 90 from values just above, 91, 90.5, 90.1 etc, y goes to - infinity.)

Sometimes it may not be necessary to do all of these because answers to some of these questions may make others obvious or the answer may be it doesn't have any.

If you remember your objective ( to know what the curve looks like and how it behaves) you can often short cut some of the process. For example, *all cubics have a similar sketch* (I don't mean similar in the scale factor sense) so you can use this and get straight to the important things like does it start in the 3rd quadrant and go to the first, or does it go from 2nd to 4th. Does it cross the x axis in one, two or three places?

Here are some you might like to try:

Have fun!

Bob

*Last edited by bob bundy (2011-02-26 22:56:13)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Au101****Member**- Registered: 2010-12-01
- Posts: 348

Thank you Bob, that was brilliant

For the first (1) I imagine that it's to do with the cubic function. i.e. an x-cubed graph can only have, at most, three roots, and you know that negative numbers cubed are negative and positive numbers cubed are positive, so the cubic curves must increase to the left, turn or inflect and then resume increasing (after a second turning point where appropriate). I probably haven't explained that very well, but I think - or at least hope - that I have the general idea. Obviously that wouldn't apply for

, though.*Last edited by Au101 (2011-02-27 00:01:10)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,787

hi Au101,

You do have the general idea but I was hoping you would practice the ideas from the list.

eg. (i) It may have a linear factor and taking that out would leave a quadratic with, at most, two more factors. So the greatest number of crossing points will be three but it remains to be shown that it will even have one.

...........................

You will be able to show this, using other things from the list.

Bob

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**Au101****Member**- Registered: 2010-12-01
- Posts: 348

Hello, bob, yes I did practise, but I just sketched the curves on paper.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,787

Ok then,

Best wishes,

Bob

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