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Graphing a cubed root function

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Sketch the curve of [itex]\sqrt[3]{(x^{2} - 1)^{2}}[/itex][/b]

    2. Relevant equations



    3. The attempt at a solution

    I determined that the domain is all real numbers, the x-int. is ±1, the y-int. is 1, the function is symmetric about the y-axis, there are no asymptotes, and here is where I get lost. I took the first derivative of the function and found it to be equal to[itex]\frac{2}{3}[/itex] (x[itex]^{2}[/itex] - 1)[itex]^{\frac{-1}{3}}[/itex] * 2x = [itex]\frac{4x}{3\sqrt[3]{x^{2}-1}}[/itex] I know one of the critical numbers is 0 and I am wondering if the other numbers is ±1.
     
    Last edited by a moderator: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2

    ehild

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    What do you mean on "the x-int. is ±1, the y-int. is 1"?

    Where is the function increasing and where is it decreasing? Try to sketch it. What happens at x=±1?

    ehild
     
  4. Nov 9, 2012 #3
    At x=±1 the function lands on the x-axis. The function is decreasing on (-∞,0) and increasing on (0,∞).
     
  5. Nov 9, 2012 #4

    ehild

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    What happens in the interval (-1,1)?

    ehild
     
  6. Nov 9, 2012 #5
    It concaves downward.
     
  7. Nov 9, 2012 #6

    SammyS

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    Since the derivative is undefined at x = ±1, those also are critical numbers.
     
  8. Nov 9, 2012 #7

    ehild

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    What is the function at x=0?
    Is it increasing or decreasing between -1 and 0, and between 0 and 1?

    ehild
     
  9. Nov 9, 2012 #8
    It is increasing between -1 and 0 and decreasing between 0 and 1.
     
  10. Nov 9, 2012 #9
    At x=0 the value is 1.
     
  11. Nov 9, 2012 #10

    ehild

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    Well, so it is decreasing in (-∞,-1), increasing between -1 and 0, decreasing between 0 and 1, increasing in (1,∞). The function is 1 at x=0 and 0 at ±1. Can you sketch it?

    ehild
     
  12. Nov 9, 2012 #11
    I have another question. I am trying to find the second derivative of the function and this is what I have. [itex] f"(x) = \frac{4*3\sqrt[3]{x^{2}- 1}- 4x*3*\frac{1}{3\sqrt[3]{x^{2}-1}}}{9\sqrt[3]{(x^{2}-1)}^{2}} = \frac{12\sqrt[3]{x^{2}-1}- 12x}{27\sqrt[3]{(x^{2}-1)^{2}}\sqrt[3]{x^{2}-1}} = \frac{12\sqrt[3]{x^{2}-1}- 12x}{27(x^{2}-1)} [/itex]. I am wondering if this is right so far.
     
  13. Nov 9, 2012 #12

    SammyS

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    Incorrect.

    The derivative of [itex]\ \ 3\sqrt[3]{x^2-1}\ \ [/itex] is [itex]\displaystyle \ \ \frac{2x}{\sqrt[3]{(x^2-1)^2}}\ .[/itex]
     
  14. Nov 9, 2012 #13
    I'm talking about the second derivative.
     
  15. Nov 9, 2012 #14

    SammyS

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    So was I.

    In taking the derivative, using the quotient rule, you took the derivative of the denominator of the first derivative incorrectly.
     
  16. Nov 9, 2012 #15
    So the equation becomes [itex] f"(x) = \frac{4*3\sqrt[3]{x^{2}-1}- 4x*\frac{2x}{\sqrt[3]{(x^{2}-1)^{2}}}}{9\sqrt[3]{(x^{2}-1)^{2}}} = \frac{12\sqrt[3]{x^{2}-1}- 8x^{2}}{9(x^{2}-1)\sqrt[3]{x^{2}-1}} = \frac{12-8x^{2}}{9(x^{2}-1)}[/itex]
     
  17. Nov 10, 2012 #16

    SammyS

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    Not quite.

    The first step is correct.
     
  18. Nov 10, 2012 #17
    So what did I do wrong.
     
  19. Nov 10, 2012 #18

    SammyS

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    [itex]\displaystyle f"(x)
    = \frac{4*3\sqrt[3]{x^{2}-1}- 4x*\frac{2x}{\sqrt[3]{(x^{2}-1)^{2}}}}{9\sqrt[3]{(x^{2}-1)^{2}}}
    = \frac{12\sqrt[3]{(x^{2}-1)^3}- 8x^{2}}{9(x^{2}-1)\sqrt[3]{x^{2}-1}}= \ \ \dots
    [/itex]
     
  20. Nov 10, 2012 #19
    Is there a problem between the second and third steps.
     
  21. Nov 10, 2012 #20
    So the cube root on the top would disappear and it would be [itex] \frac{4x^{2}-1}{9(x^{2}-1)\sqrt[3]{x^{2}-1}} [/itex]
     
    Last edited: Nov 10, 2012
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