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Math Help - Graph-min, inflection pts for a trig function

  1. #1
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    Graph-min, inflection pts for a trig function

    Usually I'm fine with most of the graphing..but the trig always trips me up..
    The problem:

    Let f be the function defined for x is greater than or equal to pi/6 and x is less than or equal to 5pi/6. F(x)= x+sin^2(x)

    a) Find all values of x for which f(x)=1

    I did it on my calculator and got x is approximately 0.6417...by hand I tried to factor w/ 0=x+sin^2-1 but that was a fail...So is my answer exact enough, or how would I get the exact answer by hand?

    b) Find the x-coordinates for all min points of f.

    So I take the f'(x), which = 1+2sin(x)cos(x), then to simply that I got it down to -1=sin(2x). What do I do from here?

    c) find the inflection pts...so I just take the second derivative right?
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  2. #2
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    Quote Originally Posted by bcahmel View Post
    Usually I'm fine with most of the graphing..but the trig always trips me up..
    The problem:

    Let f be the function defined for x is greater than or equal to pi/6 and x is less than or equal to 5pi/6. F(x)= x+sin^2(x)

    a) Find all values of x for which f(x)=1

    I did it on my calculator and got x is approximately 0.6417...by hand I tried to factor w/ 0=x+sin^2-1 but that was a fail...So is my answer exact enough, or how would I get the exact answer by hand?
    sin^2x=1-cos^2x\Rightarrow\ x+sin^2x-1=0\Rightarrow\ x-cos^2x=0\Rightarrow\ (\sqrt{x}-cosx)(\sqrt{x}+cosx)=0

    You could try the Newton-Raphson method to solve

    cos^2x-x=0

    but you will get a very close approximation.
    Last edited by Archie Meade; January 14th 2011 at 01:15 PM.
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  3. #3
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    ok thanks for help with that part...good to know I didn't learn that method in trig!
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