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Math Help - critical and inflection pts

  1. #1
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    critical and inflection pts

    find the critical points of f(x) and use the second derivative test (if possible) to determine whether each corresponds to a max or min.

    f(x)=sin(x)^2+cos(x) ,[0,pi]

    i got f'(x) and f''(x) but im just having a little trouble setting them equal to zero and solving for the critical and inflection points. Could someone help me out on that please?

    f'(x)=2*sin(x)*cos(x)-sin(x)=0 x=? -->critical pts.
    f''(x)=4*cos(x)^2-cos(x)-2=0 x=? -->inflection pts
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  2. #2
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    i got the critical pts actually, x=pi/3 and x=0
    and for the inlfection pts its cosx=2 and cosx=3/4 ..is that possible?
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  3. #3
    Super Member PaulRS's Avatar
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    For the zeros in the first derivative factorize.

    2\cdot{\sin(x)\cdot{\cos(x)}}-\sin(x)=0

    Then \sin(x)\cdot{\left(2\cdot{\cos(x)}-1\right)}=0

    From there you get \sin(x)=0 and 2\cdot{\cos(x)}-1=0

    For the second derivative

    Try this change of variable z=\cos(x) what do you see?
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  4. #4
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    i understand now, thankyou !
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