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Math Help - [SOLVED] Help! - Values

  1. #1
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    [SOLVED] Help! - Values

    I need help finding out the biggest and smallest values of;

    f(x)=2(cosx)^2 - 2sinx + 1

    without using the derivative.

    Please, need help as soon as possible!
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  2. #2
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    Quote Originally Posted by weasley74 View Post
    I need help finding out the biggest and smallest values of;

    f(x)=2(cosx)^2 - 2sinx + 1

    without using the derivative.

    Please, need help as soon as possible!
    You want values where sine is small and cosine is large for your maximum of the function. Sine is small (zero) and cosine is large (one) when x = 0, pi, 2pi, ....
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  3. #3
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    what about the smallest value..? is that just the other way around?
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  4. #4
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    Quote Originally Posted by weasley74 View Post
    what about the smallest value..? is that just the other way around?
    Correct!
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  5. #5
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    so for minimum x=pi/2 +2npi

    and for maximum x=0, x=pi+2npi ?
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  6. #6
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    Hello, weasley74!

    Find the biggest and smallest values of: . f(x)\:=\:2\cos^2\!x - 2\sin x + 1
    without using the derivative.
    Replace \cos^2\!x with 1 - \sin^2\!x

    We have: . f(x) \:=\:2(1-\sin^2\!x) - 2\sin x + 1 \quad\Rightarrow\quad f(x) \:=\:-2\sin^2\!x - 2\sin x + 3


    Let u \,=\,\sin x\!:\;\;f(u) \:=\:-2u^2 - 2u + 3

    This is a down-opening parabola; its maximum is at its vertex.
    . . The vertex is at: . u\:=\:\frac{-b}{2a} \:=\:\frac{-(-2)}{2(-2)} \:=\:-\frac{1}{2}

    Hence, the maximum is: . f\left(-\frac{1}{2}\right) \:=\:-2\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) + 3 \;=\;\boxed{\frac{7}{2}}


    Normally, there is no minimum for this parabola.
    . . However, the values of u are limited: . -1 \leq u \leq 1

    The minimum occurs at an endpoint of the interval.

    . . And we find that the minimum is: . f(1) \:=\:\boxed{-1}

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  7. #7
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    Quote Originally Posted by Soroban View Post

    This is a down-opening parabola; its maximum is at its vertex.
    . . The vertex is at: . u\:=\:\frac{-b}{2a} \:=\:\frac{-(-2)}{2(-2)} \:=\:-\frac{1}{2}

    how do you get to this point? what is a and b?
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  8. #8
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    Quote Originally Posted by weasley74 View Post
    how do you get to this point? what is a and b?
    Are you familiar with parabolas? Do you know how to find the turning point of a parabola?

    There are many ways to get the turning point - Soroban gave one which clearly you're not familiar with. What about completing the square to get turning point form? You've gotta have met that one ....?

    f(u) = -2u^2 - 2u + 3 = -2 \left( u^2 + u - \frac{3}{2} \right) = -2 \left( \left[ u + \frac{1}{2} \right]^2 - \frac{7}{4} \right) = -2\left(u + \frac{1}{2} \right)^2 + \frac{7}{2}.

    Maximum turning point at \left(-\frac{1}{2}, \, \frac{7}{2}\right), that is, f(u) has a maximum value of  \, \frac{7}{2} \, (corresponding to u = -\frac{1}{2} \Rightarrow \sin x = -\frac{1}{2} .....)
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  9. #9
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    Quote Originally Posted by colby2152 View Post
    You want values where sine is small and cosine is large for your maximum of the function. Sine is small (zero) and cosine is large (one) when x = 0, pi, 2pi, ....
    Not wanting to rub it in ....... The error in this argument is that, unfortunately, it forgets that \sin x can be negative (and hence - 2 \sin x can also make a positive contribution).
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    Not wanting to rub it in ....... The error in this argument is that, unfortunately, it forgets that \sin x can be negative (and hence - 2 \sin x can also make a positive contribution).
    Yes, that thinking only works best in the 1st quadrant.
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  11. #11
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    \begin{aligned}<br />
  f(u) &=  - 2u^2  - 2u + 3\\<br />
   &=  - \frac{{\left( {4u^2  + 4u - 6} \right)}}<br />
{2}\\<br />
   &=  - \frac{{(2u + 1)^2  - 7}}<br />
{2}.<br />
\end{aligned}

    May be this could be another way to get minimum point.
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