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Babynini

  • one year ago

Prove the equation has at least 1 real root.

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  1. Babynini
    • one year ago
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    #56 :)

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  2. ganeshie8
    • one year ago
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    Have you tried using IVT ?

  3. Babynini
    • one year ago
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    Noo. I'm not sure where to start.

  4. Babynini
    • one year ago
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    I don't know what IVT is o.o

  5. dan815
    • one year ago
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    find that the function has a positive and a negative value

  6. dan815
    • one year ago
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    or use the fact that it's an odd functions, so the ends, have one going to infnite and the other going to negative infinity

  7. dan815
    • one year ago
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    and thsi is an analytical function meaning there are no holes, and its connected so it must have passed through 0 to do this

  8. dan815
    • one year ago
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    |dw:1444528198685:dw|

  9. Babynini
    • one year ago
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    Ahh k. ^ makes sense thus far.

  10. ganeshie8
    • one year ago
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    If you're not familiar with IVT yet, you may also simply use the fundamental theorem of algebra : A polynomial of degree "n" has exactly "n" roots and out of these "n" roots, the complex roots always come in conjugate pairs.

  11. ganeshie8
    • one year ago
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    since the polynomial in question has a degree of 5, which is odd, it follows that at least one root must be real.

  12. Babynini
    • one year ago
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    degree of 5? hm?

  13. ganeshie8
    • one year ago
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    Look at the equayion in problem #56 first convinvce yourself that it is a polynomial equation

  14. dan815
    • one year ago
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    just do this f(10000) = positive f(-1000000) = negative you are done

  15. dan815
    • one year ago
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    since its a continuos function these 2 points are connected in between in some way, and the line must obviously pass y=0

  16. dan815
    • one year ago
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    the curve* must pass if u are picky -.-

  17. Babynini
    • one year ago
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    Haha k. Well, what does it mean for it to have a real root? that just means it passes through 0?

  18. dan815
    • one year ago
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    thats what a root is

  19. dan815
    • one year ago
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    a real root means u have an x value where that intersects the x axis

  20. ganeshie8
    • one year ago
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    real roots are same as x intercepts in the graph

  21. Babynini
    • one year ago
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    Ahhh k. So as soon as I recognize it's a polynomial then I basically know it has root? -.- why do I want to go about proving it then?

  22. ganeshie8
    • one year ago
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    for example, below graph has 3 x intercepts, so we say the number of real roots are 3 |dw:1444528871031:dw|

  23. dan815
    • one year ago
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    highest is odd degree* polynomial

  24. dan815
    • one year ago
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    even highest degree polynomials dont need to have real root

  25. Babynini
    • one year ago
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    I see, I see. Ok, this is making much more sense.

  26. Babynini
    • one year ago
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    So for part a, I legit just have to say: "since the polynomial in question has a degree of 5, which is odd, it follows that at least one root must be real."

  27. dan815
    • one year ago
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    you might need more support, the better way is to show that htere is a positive and a negative value

  28. dan815
    • one year ago
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    and say since its continuous.. it must pass thru xaxis

  29. dan815
    • one year ago
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    you cant just be saying oh its degree 5, so it just works, u are going to have to link polynomial functiosn to their odd function counter parts

  30. dan815
    • one year ago
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    you are going to have to state something like since its a composition of odd and even functions x^5, ax^4, bx^3,cx^3,dx,3 as x goes to infinite the additions of the other degrees will pare in comparision and the end points will exhibit the odd degree characteristics

  31. dan815
    • one year ago
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    pale*

  32. Babynini
    • one year ago
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    hm ok. Should I do any equations to prove it?

  33. dan815
    • one year ago
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    u could do x^5 + ax^4 + bx^3 + cx^3 + dx +3 =0 divide everything by x^4 x+a+b/x + c/x^2 + d/x^3 + 3/x^4=0 now u can check x=inf and -inf if u set x=inf ull end up with inf +a if u set x=-inf ull end up with -inf +a

  34. dan815
    • one year ago
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    so u have shown that this functions goes to inf and -inf at its end points

  35. Babynini
    • one year ago
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    hmm ok. Why did you add all those extra numbers? o.o

  36. dan815
    • one year ago
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    the abcd?

  37. Babynini
    • one year ago
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    Yeah. If the original is: x^5-x^2+2x+3

  38. dan815
    • one year ago
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    its just more general, u can plug in the values for a b c d respectively

  39. dan815
    • one year ago
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    i couldnt remember your question exactly so

  40. Babynini
    • one year ago
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    ^ thanks so much. I think I get it :) looking over notes and looking at all you said.

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