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iamsammybear

  • 2 years ago

Consider f(x)=sin^2(x)-cos(x) on the interval 0 is less than or equal to x which is less than or equal to pi. Determine the local extrema of f(x).

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  1. iamsammybear
    • 2 years ago
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    \[f(\theta)=\sin^{2}(\theta)-\cos(\theta)\] \[0\le \theta \le \pi\]

  2. iamsammybear
    • 2 years ago
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    I know one of the local extrema is at x=0. There are two more.

  3. niksva
    • 2 years ago
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    in order to find the local extrema first differentiate f(x) with respect to x and then put f '(x) = 0 this will help u in getting local extrema

  4. iamsammybear
    • 2 years ago
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    I get:\[\sin(\theta)(2\cos(\theta)+1)=0\]

  5. iamsammybear
    • 2 years ago
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    I know the sin (theta) piece is what give the local extrema at x=0. But I can't figure out the cos part.

  6. niksva
    • 2 years ago
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    ok 2{cos (theta) + 1} = 0 now can we write cos(theta) + 1 = 0?

  7. niksva
    • 2 years ago
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    cos(theta) = -1 cos(theta) = cos(n* pi) where n is an odd integer is this step clear to u ?

  8. niksva
    • 2 years ago
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    i m sorry its wrong

  9. iamsammybear
    • 2 years ago
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    Yes, I am trying to find the angle that gives the odd integer?

  10. niksva
    • 2 years ago
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    let me correct its 2cos(theta) +1 = 0 2cos(theta) = -1 cos(theta) = -1/2 is this step clear to u?

  11. niksva
    • 2 years ago
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    as we know cosine function is negative in second and third quadrant therefore , value of theta must lie in pi/2 < theta < 3*pi/2

  12. iamsammybear
    • 2 years ago
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    2pi/3 or 4 pi/3

  13. niksva
    • 2 years ago
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    yup

  14. iamsammybear
    • 2 years ago
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    Do I then restrict those answers based on the domain?

  15. niksva
    • 2 years ago
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    yeah

  16. iamsammybear
    • 2 years ago
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    So then 2pi/3 is the only one that works?

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