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Math Help - Solving a polynomial problem without guess & check!

  1. #1
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    Solving a polynomial problem without guess & check!

    Hey everyone,

    I have the following problem and I know I could solve with guess & check, but how would I solve it to be able to prove my answer?


    Problem:

    Find the range of values of 'c' such that the function f(x) = 2x3 - x2 - 6x + c has a zero between x = 0 and x = 1





    Thanks everyone!
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  2. #2
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    Quote Originally Posted by qcom View Post
    Hey everyone,

    I have the following problem and I know I could solve with guess & check, but how would I solve it to be able to prove my answer?

    Problem:

    Find the range of values of 'c' such that the function f(x) = 2x^3 - x^2 - 6x + c has a zero between x = 0 and x = 1
    in future, use the caret symbol ^ to indicate exponents, or learn Latex.



    f(x) = 2x^3 - x^2 - 6x + c

    f(0) = c

    f(1) = 2 - 1 - 6 + c = c - 5

    for a zero to exist between x = 0 and x = 1, there has to be a sign change of the function values

    if f(0) = c > 0 , then f(1) = c-5 < 0

    0 < c < 5


    looking at the other possibility ...

    if f(0) = c < 0 , then f(1) = c-5 > 0

    looks like that possibility can't happen ... do you see why?
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  3. #3
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    Sorry for the messy exponents, what is Latex by the way?

    Now, I understood how you logically proceeded with the problem, now I think it's not possible because the question calls for the zeros to be between 0 and 1 and that would require a sign change, yet a sign change is impossible between those two integers?
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  4. #4
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    Quote Originally Posted by qcom View Post
    Sorry for the messy exponents, what is Latex by the way?

    Now, I understood how you logically proceeded with the problem, now I think it's not possible because the question calls for the zeros to be between 0 and 1 and that would require a sign change, yet a sign change is impossible between those two integers?
    I'm afraid you don't understand ... I checked two possible cases.

    Case #1 : f(0) positive and f(1) negative. this case worked as shown below ...

    f(x) = 2x^3 - x^2 - 6x + c

    f(0) = c

    f(1) = 2 - 1 - 6 + c = c - 5

    for a zero to exist between x = 0 and x = 1, there has to be a sign change of the function values ...

    if f(0) = c > 0 , then f(1) = c-5 < 0

    0 < c < 5


    Case #2 : f(0) negative and f(1) positive does not work.


    I suggest you graph f(x) using values of c between 0 and 5 and confirm for yourself that a zero for f(x) occurs between x = 0 and x = 1.
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  5. #5
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    OK, I see where I was confused, thanks for your help and patience!
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