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Math Help - relative extrema

  1. #1
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    relative extrema

    use both the first and second derivative test to show that f(x)=sin^2 x has a relative minimum of x=0

    i keep getting that x=0 has no min or max because it equals zero in the second derivative test this is what im doing see if you can see what im doing wrong

    f '(x)=2cosx and i set that equal to zero and i get x=0 then
    f ''(x)=-sinx which is equal to zero when i plug x in and that gives me no min or max
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  2. #2
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    Quote Originally Posted by vinson24 View Post
    use both the first and second derivative test to show that f(x)=sin^2 x has a relative minimum of x=0

    i keep getting that x=0 has no min or max because it equals zero in the second derivative test this is what im doing see if you can see what im doing wrong

    f '(x)=2cosx and i set that equal to zero and i get x=0 then
    f ''(x)=-sinx which is equal to zero when i plug x in and that gives me no min or max
    You have to use the chain rule here:

    f(x)=\sin^2(x)~\implies~f'(x)=2\sin(x)\cdot \cos(x)

    Go ahead!
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