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Homework Help: Derivatives that don't exist

  1. Jan 6, 2004 #1
    What happens if you are trying a first and second derivative test on a function and they do not exist on the function (they only give errors)

    Also, is it possible that the critical values of a function (max, mins, points of inflection) don't extist on a function


    y=-4x(x^2 -6)^0.5
    y'=-8(x^2 -6)^-0.5 (x^2 -3)
    y''=8x(x^2-6)^-1.5 (x^2 -9)

    When finding the max and mins set y'=0
    and i find that the max and mins are at -root3 and root3 (i can't type a square root sign)
    but they don't exist in the function ( when i susbstitute that into the function i get error because you can't squaer root a negative number.

    What do i do?
  2. jcsd
  3. Jan 6, 2004 #2


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    First off, I'd like to point out that at √3, what you wrote for f' does not exist either. (to make the √ symbol, write "& radic ;" but without the quotes and spaces)

    Anyways, are you sure you computed the derivative correctly?
  4. Jan 6, 2004 #3
    ya, my derivative could be wrong i tried to do they quickly
    but my question is what happens to the graph when -&radic;6 < x < &radic;6
    Last edited: Jan 6, 2004
  5. Jan 6, 2004 #4


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    It doesn't exist. Nothing there.
  6. Jan 6, 2004 #5


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    Well, don't do derivatives quickly! I would like to know how you arrived at that answer. It doesn't seem to be anything like what I get.

    I get, for the derivative of y=-4x(x^2 -6)^0.5,
    y'= -4(x^2- 6)^0.5- 2x(x^2-6)^-0.5(2x)= 4(x^2-1)(x^2-6)^-0.5 which is 0 at -1 and 1 and undefined at the endpoints +/- &radic;(6).

    As for "what happens to the graph when -&radic;(6) < x < &radic;(6)" the answer is nothing happens to it. The function itself is only defined for x<-&radic;(6) and x> &radic;(6). For x between -&radic;(6) and &radic;(6), x2-6 is negative and does not return a real number.
    Last edited by a moderator: Jan 6, 2004
  7. Jan 6, 2004 #6
    I tried both derivatives again and got the same answer.

    and to find max and mins you set the first derivative equal to 0. right?

    Then you solve for x. doind that you find that when x= +/- &radic;3 the derivative =0 . so they are the maximums and minimums, but they don't exist on the graph b/c when you sub +/- &radic;3 back into the original equation you get an error b/c you can't take the square root of a negative.

    SO i'm wonder what would happen since the max and mins don't exist on a graph?
  8. Jan 6, 2004 #7
    So i guess there would just be a big gap in the middle.
    But what would you use the max and mins for?

    What are the max and mins of the graph?

    How do you find the asymptotes? do i use the max and mins that i found?
  9. Jan 7, 2004 #8


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    You did the derivative again and got the same answer? The same as you got before or the same as I did?

    I wrote that bit about the derivative being 0 at 1 and -1 BEFORE I notice that the function is not defined between -&radic;(6) and &radic(6) and then did not change it. Obviously, if the function itself is not defined at 1 and -1, then neither is the derivative!

    This function has no max or min. The graph is a curve (almost a straight line) from the upper left to (-&radic(6),0). Then there is a gap to (&radic(6),0) and a curve (almost a straight line) going off to the lower right.

    Well, not exactly. strictly speaking the point (-&radic(6),0) is a "local minimum" and (&radic(6), 0) is a "local maximum". Notice that those are the first points at which the derivative does not exist.
    Last edited by a moderator: Jan 7, 2004
  10. Jan 7, 2004 #9
    So i guess that there are no points of inflection either. right?

    How do you find pionts of inflection?
    Do you set the 2nd derivative equal to zero and solve for x?
    what do you do after that?

    No y-intercept, but there are 2 x-intercepts. right?
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