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Hello, bob, yes I did practise, but I just sketched the curves on paper.
Thank you Bob, that was brilliant
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:
Oooh that's very good, thanks again Bob
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.
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.
Has one positive and two negative roots.
Can I just sub in successive integers and show where the roots occur?
Ah that's perfect. Thank you once again Bob!
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.
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 .