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Math Help - first derivative test

  1. #1
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    first derivative test


    Hi, thank you for your help.
    i have being given this question to do, i have done the first part but am unsure about the nxt parts..
    I have y=3x^4+4x^3
    i had to find dy/dx = 12x^3+12x^2
    Then i need to show that the graph of the function y has stationary points at x=0 and x=-1, and find their co ords.
    Then i have to find if these points are min max or stationary, which i know how to do by the gradient function(the table thing not sure of its name)

    If you can explain how to do the second part that would be great. thank you
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  2. #2
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    A function f(x) has extremum points at the roots of f'(x) = 0.

    f(x) = 3x^4 + 4x^3

    f'(x) = 12x^3 + 12x^2

    Let's find f'(x) = 0,

    12x^3 + 12x^2 = 0

    \underbrace{(x^2)}_{I}\underbrace{(12x+12)}_{II} = 0

    \text{I.}~~x^2 = 0
    x = 0

    \text{II.}~~12x+12 = 0
    x = -1

    So x=\{-1,0\}

    Now use that table thing (hehe) or the second derivative test to determine whether these are minimums or maximums
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  3. #3
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    Hi, iv done a table for x=-3,-2,-1,0,2,and 4
    and got dy/dx = -216, -48, 0 , 0, 144 and 960. But im not sure which of these are max and mins?

    Also by oing this how does it answer the q, determine wether the stationary points are max or min? as -1 and 0 are statinary arent they?

    :S thanks
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  4. #4
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    Quote Originally Posted by Chez_ View Post
    Hi, iv done a table for x=-3,-2,-1,0,2,and 4
    and got dy/dx = -216, -48, 0 , 0, 144 and 960. But im not sure which of these are max and mins?

    Also by oing this how does it answer the q, determine wether the stationary points are max or min? as -1 and 0 are statinary arent they?

    :S thanks
    You look at the sign of the derivative of either side of the stationary point. That's the reson for setting up the table.

    I'll do x = -1:

    f'(x) = 12x^3 + 12x^2 = 12x^2 (x + 1)

    To the left of x = -1: x = -2 and f'(-2) = (48)(-3) < 0.

    To the right of x = -1: x = -1/2 and f'(-1/2) = 3(1/2) > 0.
    (Note that using x = 0 as the value of x to the right is NO GOOD because x = 0 is a stationary point).

    Therefore by the sign test, x = -1 is minimum turning point.

    Unfortunately, your table had no useful value of the derivative to right of x = -1 (and indeed to the left of x = 0). You needed something like the value of f'(-1/2). That's why you ran into trouble.

    To test the nature of the stationary point at x = 0, I'd suggest looking at the sign of the derivative at x = -1/2 (and you've already got that from what I did above) and x = 2 (straight from your table). By the way ..... x = 1 is a nice simple value .... Why didn't you use it?

    (For confirmation purposes: There's a stationary point of inflexion at x = 0).

    Oh yeah ...... one small point:

    The value of f'(-2) is NOT -48 .... Not that it really matters in this case - the important thing is what the sign of the value is, not the actual value itself .....
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  5. #5
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    Quote Originally Posted by wingless View Post
    [snip]
    So x=\{-1,0\}

    Now use that table thing (hehe) or the second derivative test to determine whether these are minimums or maximums
    Note: The second derivative test fails for x = 0 since f''(0) = 0 ......

    The sign test has to be used to test x = 0. And the calculated table of values is no use in applying this test for the reason explained in my above reply.
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Note: The second derivative test fails for x = 0 since f''(0) = 0 ......

    The sign test has to be used to test x = 0. And the calculated table of values is no use in applying this test for the reason explained in my above reply.
    Aaah that's not fair.. I really don't like to make these tables..

    <br />
          \begin{array}{c|ccccccccc}<br />
x&-2&&-1&&-0.5&&0&&1           \\\hline<br />
f'(x)&-48&\nearrow&0&\nearrow&1.5&\searrow&0&\nearrow&24<br />
          \end{array}<br />

    f'(x) always increases around x=-1. So this point is a minimum.

    f'(x) decreases on the left of x=-1, but on the right it starts increasing.

    We can check the results with the graph,

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  7. #7
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    Hi, thank you all for your help, i have written it down and worked it out, but i just want to check that i have written it down right and actually answered the question

    i followed what wiggles said, to find the stationary points are -1 and 0. is that right?

    i have then looked mr fantastics explanation, to determin weher these points were mins or max or statinary. But i dont quite follow where you have got the x=-2 then the working after from ..and if the answer to this is <0 y does that show that -1 is the min? Also by saying that x=0 is stationary as the points on either side of it are decreasing and increasing but the gradient of x=0 remains the same, would that be ok?
    I have managed the graph also,
    Thank you!!!
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