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    2. Use the method of Interval Bisection twice to get approximate solutions to these equations, starting with the given intervals:

    a) sin 3x = x2 - 1, -1 < x < 0

    answer
    2.

    a) f(-0.25) = 0.256



    ....??????
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    f(x) = -x^2 + sin3x + 1
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    (Original post by ckfeister)
    f(x) = -x^2 + sin3x + 1
    f(-0.5) \approx -0.2474 \not \approx 0.256

    f(-0.25) \approx 0.256 \not \approx 0.38
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    (Original post by RDKGames)
    f(-0.5) \approx -0.2474 \not \approx 0.256

    f(-0.25) \approx 0.256 \not \approx 0.38
    How...
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    (Original post by ckfeister)
    How...
    Plug it into your calculator correctly...
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    (Original post by RDKGames)
    Plug it into your calculator correctly...
    I got the plus and minus at -x^2 confused, I did -0.5^2 not -(-0.5^2)
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    (Original post by ckfeister)
    I got the plus and minus at -x^2 confused, I did -0.5^2 not -(-0.5^2)
    Should be -(-0.5^2), something to look out for more carefully in the future
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    (Original post by ckfeister)
    I got the plus and minus at -x^2 confused, I did -0.5^2 not -(-0.5^2)
    Are you using degrees or radians? In general you use radians with this sort of question
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    (Original post by alfredholmes)
    Are you using degrees or radians? In general you use radians with this sort of question
    Radians, I've done it now.
 
 
 
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