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Graph sketching problem

  1. Nov 24, 2013 #1
    1. The problem statement, all variables and given/known data
    I've been looking at the problem for a while and I just can't figure it out. It is graph sketching.

    [STRIKE]y=x2(x2-4)2[/STRIKE]
    the correct equation is this:
    y=x2(x-4)2


    2. Relevant equations

    Solve for critical points, and inflection points in order to get points needed to graph function.

    3. The attempt at a solution

    [STRIKE]y=x2(x-4)2[/STRIKE]
    y=x2(x2-4)2
    y=x4-8x3+16x2

    y'=4x3-24x2+32x
    0=4x(x-4)(x-2)
    x=0,2,4

    local min: 0,4
    local max: 2

    y"=12x2-48x+32

    I don't know what to do to solve for the inflection points.
     
    Last edited: Nov 24, 2013
  2. jcsd
  3. Nov 24, 2013 #2
    Well, inflection points can be arrived at by solving f "(x)=0 and finding values of x and substituting them again in the given equation.
    That should give you all inflection points of that function.
     
  4. Nov 24, 2013 #3
    Yeah I'm aware of that, and I've tried, but I can't figure out how to solve for x.
     
  5. Nov 24, 2013 #4

    FeDeX_LaTeX

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    Are you asking us how you might solve ##12x^2 - 48x + 32 = 0##?
     
  6. Nov 24, 2013 #5
    Yes, or at least some pointers in the right direction. I'd like to try to do it myself if possible.
     
  7. Nov 24, 2013 #6

    FeDeX_LaTeX

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    Have you seen the quadratic formula before?

    In relation to sketching the original curve, it might save you some time to note whether or not the function is odd or even.
     
  8. Nov 24, 2013 #7
    Yes! I didn't even think about that! Thank you for your help! It should be an even function just by looking at the formula. How do you figure out algebraically if it is even or odd?
     
  9. Nov 24, 2013 #8

    FeDeX_LaTeX

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    In general, a function ##f(x)## is odd if ##f(-x) = -f(x)##, and even if ##f(-x) = f(x)##. (You can also decompose a function into an even part and an odd part, but that's irrelevant here.)

    What happens when you replace ##x## with ##-x## in your function?
     
  10. Nov 24, 2013 #9
    f(x) = x4-8x3+16x2

    f(-x) = (-x)4 -8(-x)3+16(x)2

    = x4 + 8x3 + 16x2

    So I was wrong. It's actually neither even or odd.
     
  11. Nov 24, 2013 #10

    SteamKing

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    From the OP:
    y = x[itex]^{2}[/itex](x[itex]^{2}[/itex]-4)[itex]^{2}[/itex]
    y = x[itex]^{4}[/itex]-8x[itex]^{3}[/itex]+16x[itex]^{2}[/itex]

    You need to work on your basic algebra skills: the first equation does not lead to the second.

    y = x[itex]^{2}[/itex](x[itex]^{2}[/itex]-4)[itex]^{2}[/itex]
    y = x[itex]^{2}[/itex](x[itex]^{4}[/itex]-8x[itex]^{2}[/itex]+16)
    y = (x[itex]^{6}[/itex]-8x[itex]^{4}[/itex]+16x[itex]^{2}[/itex])
     
  12. Nov 24, 2013 #11
    Oh god. Well there's my issue. I forgot to foil. Thanks for your help. I don't think I'll have an issue with this problem anymore. That's embarrassing.
     
  13. Nov 24, 2013 #12

    Ray Vickson

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    The function is even; x^2 is even and x^2 - 4 is even, so x^2 * (x^2-4)^2 is even. No work needed!
     
  14. Nov 24, 2013 #13
    I forgot to foil so I came out with the wrong function.
     
  15. Nov 24, 2013 #14

    FeDeX_LaTeX

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    Yes, but you can tell that the function is even without any expansion, since you have a product of two even functions.
     
  16. Nov 24, 2013 #15
    Yeah I know. I said above that it should be even. However when I actually did it, I used the wrong function, which was the function I for some reason didn't foil. I appreciate everyone's help.
     
  17. Nov 24, 2013 #16
    This is going to sound funny, but I typed in the function above wrong. I didn't sleep much last night as you can probably tell.

    y=x2(x2-4)2

    It is supposed to be y=x2(x-4)2

    So I have the same problem. I'll try doing the quadratic formula to see what I get.
     
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