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IsTim

  • 4 years ago

how to graph g(x)=3x^4=3x^2

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  1. IsTim
    • 4 years ago
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    I'm looking thru my old notes now, but I can't find anything that could help.

  2. IsTim
    • 4 years ago
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    I simplified the equation so: g(x)=3x^2(x^2+1)

  3. IsTim
    • 4 years ago
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    I was thinking apq, but I don't know if that applies to this.

  4. sasogeek
    • 4 years ago
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    why do u have 2 equal signs to one function?

  5. IsTim
    • 4 years ago
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    Oops.

  6. IsTim
    • 4 years ago
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    g(x)=3x^4-3x^2 My bad.

  7. IsTim
    • 4 years ago
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    \[g(x)=3x ^{4}-3x ^{2}\] If you want a cleaner version.

  8. anonymous
    • 4 years ago
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    use a graphing calc

  9. anonymous
    • 4 years ago
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    3x^2(x^2-1) you plug in points frankly :-/ it looks like a x^2 graph.

  10. IsTim
    • 4 years ago
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    So I just plug in values? That's feels "brute". There's no other way?

  11. IsTim
    • 4 years ago
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    @ Mario; I'm studying for an exam. I don't get those.

  12. anonymous
    • 4 years ago
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    we get to use graph calcs on my exams

  13. anonymous
    • 4 years ago
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    plug in pounts. their are zeroes at 0, 1,-1.

  14. anonymous
    • 4 years ago
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    the rest is just brute force plugging. yes. that's how it works.

  15. IsTim
    • 4 years ago
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    Oh well. I was looking for some equation rearranging. Would that work?

  16. IsTim
    • 4 years ago
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    Lucky Mario.

  17. anonymous
    • 4 years ago
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    yet i still did poorly lol

  18. JamesJ
    • 4 years ago
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  19. IsTim
    • 4 years ago
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    IF possible, please give an explaination of how to dervie the graph from teh equatioon.

  20. JamesJ
    • 4 years ago
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    Ok. If g(x)=3x^4-3x^2 first we'd like to know the zero; i.e., the intercepts on the x-axis. Setting g(x) = 0, we have \[ 3x^2(x^2 - 1) = 0 \] Hence \[ x = 0, \pm 1 \] Next, what's the y-intercept: y = g(0) = 0. Next, critical values ...

  21. JamesJ
    • 4 years ago
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    \[ g'(x) = 12x^3 - 6x = 0 \] if and only if \[ 6x(2x^2 - 1) = 0 \] i.e., \[ x = 0, \pm 1/\sqrt{2} \] The second derivative is \[ g''(x) = 36x^2 - 6 = 6(6x^2 - 1) \] ...

  22. JamesJ
    • 4 years ago
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    It's not hard now to show that x = 0 must be a local max and \[ x = \pm 1/\sqrt{2} \] are local mins. So now we have the behavior of the function in the interval [-1,1] What happens outside that?

  23. JamesJ
    • 4 years ago
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    The next thing we observe is that g(x) is an even function, g(-x) = g(x) meaning the function is symmetric about the y-axis. As \[ g(x) = 3x^2(x^2 - 1) \] it is clear that for \( x > 1 \), \( g(x) > 0 \) and as \( x \rightarrow \infty \), \( g(x) \rightarrow \infty \). We now have all the information we need to draw the graph and it's consistent with the picture I posted above. Make sense?

  24. IsTim
    • 4 years ago
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    It's higher level information that I don't understand, but I think if I read thru it a bit more, I'll understand. Thank you very much.

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