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  1. #1
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    Calculus: Curve Sketching

    I have a calculus test tomorrow on curve sketching. I have to know how to find the asymptotes, max/min pts, pt of inflection and then sketch it...

    Im having abit of trouble with max/min pts. I'd really appreciate it if anyone could help me with this.

    Lets do a short example. Say x^4 - 8x^3 + 18x^2..

    so the first step would be to find dy/dx

    dy/dx = 4x^3 - 24x^2 + 36x

    = 4x(x^2 - 6x +9)

    =4x(x - 3) ^2

    Set dy/dx = 0
    0 = 4x(x - 3) ^2
    x = 0

    x = 3

    Now this is where im stuck at.. what would be my next step?

    (Note: Also i have seen some questions being solved by finding the second derivative... Could you also explain to me which should be used, first or second derivative, when it comes to determining the max/min pts)

    Thank you so much for any help you guys would provide ^_^

  2. #2
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    Ok.
    To find the minimums/maximus, you need to find where the FIRST derivative equals cero. Then, find the SECOND derivative, and if it's positive on those values, it means it's a minimum, and if it's negative, it means it's a maximum. If it's 0, then it's an inflection point.

    So, in your example, x=0 and x=3
    Then derivate again, d(4x^3 - 24x^2 + 36x)/dx=12x^2-48x+36
    Evaluate in 0 and 3, it gives 36 and 0 respectively.

    So in 0, it's a minimum, in 3 it's an inflection point.

    That's it

    If I could draw, I could explain a lot better, but ok...

  3. #3
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    It'll be faster for me to just solve the problem, explaining the method I use than anything else... so here goes.

    For the function f(x) = x^4 - 8x^3 + 18x^2, find:
    (a) Find all critical points, and classify them.
    (b) Find all points of inflection
    (c) Assymptotes
    (d) Sketch a curve of the function

    -----------------------------

    Preliminary work:

    f(x) = x^4 - 8x^3 + 18x^2
    f'(x) = 4*x^3 - 24*x^2 + 36*x
    f''(x) = 12*x^2 - 48*x + 36

    ------------------------------

    (a) Critical points, are all the points where the first derivative is equal to zero or where the function is not differentiable (note, since this is a polynomial... it is differentiable for any value of x). These are:

    f'(x) = 4*x^3 - 24*x^2 + 36*x = 0
    4*x*(x-3)^2 = 0
    x = 0 or x = 3

    Therefore you have critical points when x = 0, and when x = 3... plugging these values of x back into the original function...

    f(x) = x^4 - 8x^3 + 18x^2

    f(0) = (0)^4 - 8*(0)^3 + 18*(0)^2 = 0
    f(3) = (3)^4 - 8*(3)^3 + 18*(3)^2 = 27

    So at (0,0) and (3,27) you have critical points. But how to classify them...

    You can do this in one of three ways. The first is by simply plugging in values into the function. Take a number slightly less than and slightly greater than your critical points, and compare them to your critical point.)

    So f(-1) = f(0) = (-1)^4 - 8*(-1)^3 + 18*(-1)^2 = 27
    and f(1) = (1)^4 - 8*(1)^3 + 18*(1)^2 = 11

    (-1,27) and (1,11) are therefore points on our function... and (0,0) is the critical point between them. Since 27 > 0 and 11 > 0 the point (0,0) is a local minimum.

    Doing this same test on (3,27) shows...

    f(1) = 11
    f(4) = (4)^4 - 8*(4)^3 + 18*(4)^2 = 32

    (1,11) and (4,32) are therefore points on the function... and (3,27) is the critical point between them. Since 11 < 27 < 32 the point (3,27) is neither a local minimum nor a local maximum. It is just a stationary point.

    THE SECOND method of testing the critical points is similar to the first but uses the first derivative. Take a point slightly to the left and slightly to the right of your critical point. And plug it into the function for the first derivative...

    f'(x) = 4*x^3 - 24*x^2 + 36*x

    f'(-1) = 4*(-1)^3 - 24*(-1)^2 + 36*(-1) = -64
    f'(1) = 4*(1)^3 - 24*(1)^2 + 36*(1) = 16
    f'(-1) < 0 and f'(1) > 0 and (0,0) is the critical point between them. Therefore it is a local minimum. (Function decreases to get to it, and then increases afterwards.)

    f'(1) = 4*(1)^3 - 24*(1)^2 + 36*(1) = 16
    f'(4) = 4*(4)^3 - 24*(4)^2 + 36*(4) = 16
    f'(1) > 0 and f'(4) > 0 and (3,27) is the critical point between them. Therefore it is an inflection point. (Function is changing in the same way... in this case increasing... on both sides of it.)

    The FINAL method of testing the critical points... is to use the second derivative test.

    f''(x) = 12*x^2 - 48*x + 36

    f''(0) = 12*(0)^2 - 48*(0) + 36 = 36
    Since 36 > 0 the point (0,0) is a local minimum.

    f''(3) = 12*(3)^2 - 48*(3) + 36 = 0
    Because the second derivative test gives a result of 0, no conclusions can be drawn. (Hence, why I perfer the other methods... )

    ----------------------------------------

    (b) Points of inflection has to do with the curvature of a function. But for your case, just look for all the points where the second derivative is equal to zero...

    f''(x) = 12*x^2 - 48*x + 36 = 0
    a = 12, b = -48, c = 36

    x = (-b +/- sqrt(b^2 - 4ac))/(2a)
    = (-(-48) +/- sqrt((-48)^2 - 4(12)(36)))/(2(12))
    = (48 +/- sqrt(2304 - 1728)) / (24)
    = 2 +/- (sqrt(576) / 24)
    = 2 +/- (24 / 24)
    = 2 +/- 1
    x = 1 or x = 3

    Therefore you have points of inflection when x = 1 and when x = 3 plugging these values into the function

    f(x) = x^4 - 8x^3 + 18x^2

    f(1) = (1)^4 - 8*(1)^3 + 18*(1)^2 = 11
    f(3) = (3)^4 - 8*(3)^3 + 18*(3)^2 = 27

    so (1,11) and (3,27) are points of inflection.

    -------------------------------------

    (c) To find the horizontal assymptotes, take the limits as x --> infinity and as x --> -infinity for the original function: f(x) = x^4 - 8x^3 + 18x^2

    The limit as x approaches inifinity is infinity
    The limit as x approaches -infinity is infinity.

    Polynomials have no vertical assymptotes because they are defined for every real x.

    --------------------------------------

    (d) I won't do curve sketing for you, but here is the general premise...

    Start by finding the y-intercept of your function.
    f(x) = x^4 - 8x^3 + 18x^2
    f(0) = (0)^4 - 8*(0)^3 + 18*(0)^2 = 0
    So (0,0) is your y intercept.

    Then factor your function to find the x-intercepts.
    f(x) = x^4 - 8x^3 + 18x^2
    f(x) = x^2 * (x^2 - 8*x + 18)
    x^2 = 0, when x = 0.
    x^2 - 8*x + 18 = 0 when you are using complex numbers, but never for real numbers.

    Then put all your critical points and inflection points on your graph. You already have (0,0) and the second critical point is (3,27). (Recall (0,0) is a local minimum and (3,27) is an inflection point.) (1,11) is your other inflection point.

    If you want to make your sketch even more accurate add the other points you solved for while doing work on the problem. (-1,27) and (4,32) were the ones we used. Also remember that as x --> +/- infinity the function is going toward infinity.

    -------------------------------------

    We can also say at this time that because (0,0) is the only local minimum. AND because the function is going toward infinity at both ends, therefore the point is an absolute minimum for the function.

  4. #4
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    LOL well if you say thats faster demondo

    once youve equated your derivative to 0 you can solve for x to get a turning point
    we dont know the nature of this turning point until we take the second derivative

    you differentiate it again, then plug in your values for x
    if the result is > 0 then you have a minimum, if its < 0 then its a maximum

    a point of inflexion is when your second derivative is 0, and so is your third
    - Simon


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  5. #5
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    Don't worry Yazan, it's a lot lees complicated than it looks like, once you get a real explanation for it

  6. #6
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    WOW.. demondo, Xorlium, Nem. Thank you so much for your help. This all makes much more sence now..

    Now, all i have to do is simply remember all these steps.. Its quite alot of steps.. however seems simple enough to get a good mark on this test..

    Thanks again for your hlep guys. Im going to try a few problems, and see how it goes. If i need any help, i'll let u guys know. Thank
    Last edited by DrVirus; 01-23-2005 at 05:50 PM.

  7. #7
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    Nem: if the second derivative test gives a result of zero, the test FAILS and you cannot make any conclusions about the result. For DrVirus's case you can assume it was a point of inflection if the second derivative gives a result of zero. HOWEVER, that is not always the case... In order for a point to be an inflection point, at the very least, these two conditions must be met.

    1) The second derivative test gives a result of zero
    2) The concavity changes on either side of the inflection point.

  8. #8
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    Hm. And me was here thinking that the definition of inflection point was "when the concavity changes"...

  9. #9
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    the way my teacher explains it is through tis way:

    "When you graph changes from a happy face () to a sad face () or from a sad face () to a happy face ()"

  10. #10
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    lol calc in grade 12 is ultra easy. actually its easy in 110 too. i got 80 and i never did a single assignemtn and studied very little for my tests. I think the longest studying i did was when i asked afew questions and had them answered here.

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