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Thread: [SOLVED] Trignometry Help!

  1. #1
    Member looi76's Avatar
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    [SOLVED] Trignometry Help!

    Question:
    Find the maximum value and minimum value of each of the following functions. In each case, give the least positive value of x at which they occur.

    i. $\displaystyle 9 + \sin(4x-20)^o$

    ii. $\displaystyle \frac{30}{11 - 5\cos(\frac{1}{2}x-45)^o}$

    Attempt:

    i. Maximum Value $\displaystyle = 9+1 = 10$ , Minimum Value $\displaystyle = 9-1=8$

    ii. Maximum Value = $\displaystyle \frac{30}{11-5\times1} = 5$ , Minimum Value = $\displaystyle \frac{30}{11-5\times-1} = 1.875$

    How can I get the values of x in both questions?
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:
    Find the maximum value and minimum value of each of the following functions. In each case, give the least positive value of x at which they occur.

    i. $\displaystyle 9 + \sin(4x-20)^o$

    ii. $\displaystyle \frac{30}{11 - 5\cos(\frac{1}{2}x-45)^o}$

    Attempt:

    i. Maximum Value $\displaystyle = 9+1 = 10$ , Minimum Value $\displaystyle = 9-1=8$

    ii. Maximum Value = $\displaystyle \frac{30}{11-5\times1} = 5$ , Minimum Value = $\displaystyle \frac{30}{11-5\times-1} = 1.875$

    How can I get the values of x in both questions?
    You just need to solve the trig sections for +/-1

    ie
    $\displaystyle 1 = \sin(4x-20)$

    $\displaystyle -1 = \sin(4x-20)$

    $\displaystyle 1= \cos(\frac{1}{2}x-45)$

    $\displaystyle -1= \cos(\frac{1}{2}x-45)$


    Now there are multiple solutions to these, but choose the smallest positive one. It should be clear that when you solve, for example, the first one, that sin(4x-20) will then be equal to one, it's largest possible y-value. And this is corresponds to the "1" you substituted into your equation in the solution you have already provided (9 + 1 = 10)
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  3. #3
    Member looi76's Avatar
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    i. Maximum Value

    $\displaystyle Sin(4x-20) = 1$
    $\displaystyle 4x - 20 = Sin^{-1}(1)$
    $\displaystyle 4x - 20 = 90$
    $\displaystyle 4x = 90 + 20$
    $\displaystyle 4x = 110$
    $\displaystyle x = \frac{110}{4}$
    $\displaystyle x = 27.5$

    Minimum Value

    $\displaystyle Sin(4x-20) = -1$
    $\displaystyle 4x - 20 = Sin^{-1}(-1)$
    $\displaystyle 4x - 20 = -90$
    $\displaystyle 4x = -90 + 20$
    $\displaystyle 4x = -70$
    $\displaystyle x = \frac{-70}{4}$
    $\displaystyle x = -17.5$

    Why is the minimum value is wrong?
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by looi76 View Post
    i. Maximum Value

    $\displaystyle Sin(4x-20) = 1$
    $\displaystyle 4x - 20 = Sin^{-1}(1)$
    $\displaystyle 4x - 20 = 90$
    $\displaystyle 4x = 90 + 20$
    $\displaystyle 4x = 110$
    $\displaystyle x = \frac{110}{4}$
    $\displaystyle x = 27.5$

    Minimum Value

    $\displaystyle Sin(4x-20) = -1$
    $\displaystyle 4x - 20 = Sin^{-1}(-1)$
    $\displaystyle 4x - 20 = -90$
    $\displaystyle 4x = -90 + 20$
    $\displaystyle 4x = -70$
    $\displaystyle x = \frac{-70}{4}$
    $\displaystyle x = -17.5$

    Why is the minimum value is wrong?
    You need to use the positive value, ie:

    $\displaystyle Sin(4x-20) = -1$
    $\displaystyle 4x - 20 = 270$

    Notice that $\displaystyle sin(270^o) = sin(-90^o)$

    But your question specifically said "In each case, give the least positive value of x at which they occur."
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