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Math Help - First derivative test?

  1. #1
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    First derivative test?

    Consider the trigonometric function on the interval (0,2pi). Find the open intervals on which the function is increasing or decreasing.

    My first questions is

    1.f(x)=sin^2(x)+sin(x)

    2 sin(x) cos(x)+cos(x)

    cos(x) 2sin(x)+1=0 (factor)
    sin(x)=-1/2
    x=7pi/6,11pi/6 Did I do this correctly


    My second question is

    f(x)=(x-1)^(2/3)

    (2/3)(x-1)^(-1/3)(1)=0

    How would I solve this one I seem to be getting an incorrect answer
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  2. #2
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    Re: First derivative test?

    Quote Originally Posted by homeylova223 View Post

    My first questions is

    1.f(x)=sin^2(x)+sin(x)

    2 sin(x) cos(x)+cos(x)

    cos(x) 2sin(x)+1=0 (factor)
    sin(x)=-1/2
    x=7pi/6,11pi/6 Did I do this correctly
    you also need to find where \cos x= 0

    And then how will y know where the function is increasing/decreasing?
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  3. #3
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    Re: First derivative test?

    Well I can graph on my calculator and look knowing the critical numbers.
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    Re: First derivative test?

    For the second one, there are no x values wherefore f'(x)=0.
    Last edited by Siron; July 24th 2011 at 02:14 PM.
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    Re: First derivative test?

    Quote Originally Posted by homeylova223 View Post
    Well I can graph on my calculator and look knowing the critical numbers.
    Sounds good. Technology will save you a lot of time, with the second question

    Solving \displaystyle \frac{2}{3\sqrt[3]{x-1}} = 0 \implies 2=0

    which makes no sense of course, therefore has no stationary point. Just graph f(x) to see this.
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  6. #6
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    Re: First derivative test?

    To look where the functions increases/decreases, you can also make a sign table.
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    Re: First derivative test?

    Quote Originally Posted by Siron View Post
    To look where the functions increases/decreases, you can also make a sign table.
    True, this is a good idea, although I thought that was what the OP implied in post #3

    Most calculators these days have a table of values to follow.
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: First derivative test?

    Quote Originally Posted by pickslides View Post
    True, this is a good idea, although I thought that was what the OP implied in post #3

    Most calculators these days have a table of values to follow.
    Indeed, I agree, a calculator will save a lot work.
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  9. #9
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    Re: First derivative test?

    Quote Originally Posted by homeylova223 View Post

    My second question is

    f(x)=(x-1)^(2/3)

    (2/3)(x-1)^(-1/3)(1)=0

    How would I solve this one I seem to be getting an incorrect answer

    critical values of a function occur where the function's derivative equals zero or is undefined at x-values in the function's domain.

    the whole point here is that if a function has extrema on an open interval, then they occur at critical values. note that this is not a two-way street ... a function that has critical values may not have extrema located there.

    the first derivative test procedure is to check the sign of the derivative on either side of the critical value. if the sign of f' changes from positive to negative, then the original function has a maximum there ... if f' changes from negative to positive, then the original function has a minimum there. sometimes the sign will not change at a critical value ... no extrema at that value.


    your second function is an example of an extremum located where f'(x) is undefined ...

    f(x) = (x-1)^{\frac{2}{3}}

    f'(x) = \frac{2}{3(x-1)^{\frac{1}{3}}}

    note f(1) = 0 ... the function is defined.

    also note that f'(x) \ne 0 for any value of x , but f'(x) is undefined at x = 1.

    for x < 1 , f'(x) < 0 ... f(x) is decreasing

    for x > 1 , f'(x) > 0 ... f(x) is increasing

    f(x) has a minimum at x = 1 ... see the graph for confirmation.

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