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Math Help - Suck on maximum and minimum values

  1. #1
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    Suck on maximum and minimum values

    Hello. I am stuck at a point on this question.

    Identify the maximum and minimum values of the function y = 10 cos x in the interval [-2π, 2π].

    This is what I have so far (not much I know)

    10 cos [-
    2π]=10
    10 cos [
    2π]=10

    f(x)=10 sin (x)
    0=10 sin (x)
    0=sin (x)

    Nothing like the examples I was given in class so now I can't figure out what to do. Is anyone able to explain this to me please?

    Thanks everybody.
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  2. #2
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    Re: Suck on maximum and minimum values

    Quote Originally Posted by RiderMind View Post
    Hello. I am stuck at a point on this question.

    Identify the maximum and minimum values of the function y = 10 cos x in the interval [-2π, 2π].

    This is what I have so far (not much I know)

    10 cos [-
    2π]=10
    10 cos [
    2π]=10

    f(x)=10 sin (x)
    0=10 sin (x)
    0=sin (x)

    Nothing like the examples I was given in class so now I can't figure out what to do. Is anyone able to explain this to me please?

    Thanks everybody.
    Where did you get the function \displaystyle \begin{align*} f(x) = 10\sin{x} \end{align*} from? If you are trying to differentiate it, first of all, the derivative should be written as \displaystyle \begin{align*} y'(x) \end{align*} or \displaystyle \begin{align*} \frac{dy}{dx} \end{align*}, and second, the derivative of \displaystyle \begin{align*} \cos{x} \end{align*} is \displaystyle \begin{align*} -\sin{x} \end{align*}...
    Thanks from RiderMind
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  3. #3
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    Re: Suck on maximum and minimum values

    Quote Originally Posted by RiderMind View Post
    Identify the maximum and minimum values of the function y = 10 cos x in the interval [-2π, 2π].
    you only need a small bit of thought to answer this question (i.e. , no calculation is required).

    the maximum value of y is 10 and the minimum value is -10 ... now, think about why.
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  4. #4
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    Re: Suck on maximum and minimum values

    Hint: -1 \le \cos x \le 1 along the given interval. This alone should immediately lead to the solution.
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  5. #5
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    Re: Suck on maximum and minimum values

    So since -2pi would be the low end and 2pi would be the high end, and plugging those into the equation always equals 10, the minimum is -10 and maximum would be 10? Is that because cos 2pi and -2pi always equals 1?
    Last edited by RiderMind; October 4th 2012 at 06:07 PM.
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  6. #6
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    Re: Suck on maximum and minimum values

    Quote Originally Posted by RiderMind View Post
    So since -2pi would be the low end and 2pi would be the high end, and plugging those into the equation always equals 10, the minimum is -10 and maximum would be 10? Is that because cos 2pi and -2pi always equals 1?
    Sure. More specifically, \cos (-2\pi) = \cos 0 = \cos (2\pi) = 1, and \cos (-\pi) = \cos(\pi) = -1.
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  7. #7
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    Re: Suck on maximum and minimum values

    I see.

    The book wants me to explain my answer using transformations. So without using a calculator, could I say that since cos(2pi) is 1, any transformation of interval [-2π, 2π] would yield a stretch of or shrink of 10? Or am I going a little outbound on saying that?
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  8. #8
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    Re: Suck on maximum and minimum values

    the transformation was a change in amplitude ... research that term w/r to the graphs of the sine and cosine functions.
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  9. #9
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    Re: Suck on maximum and minimum values

    Hello, RiderMind

    I'm not sure what "identify" entails . . .


    \text{Identify the maximum and minimum values of }f(x) = 10\cos x\text{ in the interval }[\text{-}2\pi,\,2\pi]

    You are expected to know that: |\cos x| \,\le\,1\,\text{ and }\,\cos(n\pi) \:=\:\pm1


    So we have this list:

    . . \begin{array}{ccc} f(\text{-}2\pi) &=& +10 \\ f(\text{-}\pi) &=& -10 \\ f(0) &=& +10 \\ f(\pi) &=& -10 \\ f(2\pi) &=& +10\end{array}
    Last edited by Soroban; October 5th 2012 at 04:15 AM.
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