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Thread: Functions

  1. #1
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    Functions

    A function
    $\displaystyle
    y = f(x)
    $
    under what geometrical conditions would
    $\displaystyle
    f(x) = f^{-1} (x) ?
    $
    and whats an example of this function for which
    $\displaystyle
    f(x) = f^{-1} (x) ?
    $

    &

    does
    $\displaystyle
    \int^{1}_{-1} \sqrt\(4-x^{2}\)
    $

    $\displaystyle
    = 2{\pi}
    $
    Last edited by nath_quam; Jul 2nd 2006 at 05:37 PM.
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  2. #2
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    Quote Originally Posted by nath_quam
    A function
    $\displaystyle
    y = f(x)
    $
    under what geometrical conditions would
    $\displaystyle
    f(x) = f^{-1} (x) ?
    $
    For example, the identity function,
    $\displaystyle f(x)=x$

    ---
    Quick I am proud that yiu deleted your post
    (I am moderator I can see everything you do, in case you are wondering).
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  3. #3
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    Quote Originally Posted by nath_quam
    $\displaystyle
    \int^{1}_{-1} \sqrt\(4-x^{2}\)
    $

    $\displaystyle
    = 2{\pi}
    $
    Yes, because,
    $\displaystyle y=\sqrt{4-x^2}$ is a semi-circle.
    From $\displaystyle -2\leq x\leq 2$ the area below is,
    $\displaystyle \frac{1}{2}\pi (2)^2=2\pi$
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  4. #4
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    Hello, Nath!

    Given a function: $\displaystyle y = f(x)$
    (a) under what geometrical conditions would $\displaystyle f(x) = f^{-1}(x)\;?$

    (b) What's an example of this function for which: $\displaystyle f(x) = f^{-1} (x)\:?$

    (a) The graph must be symmetric to the line $\displaystyle y = x$.

    (b) A straight line like: .$\displaystyle y \;= \; 3 - x$
    . . .A hyperbola like: .$\displaystyle y \:= \:\frac{1}{x}$

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  5. #5
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    Would someone be able to solve this

    $\displaystyle
    \int^{1}_{-1}\sqrt\(4-x^{2}\)
    $
    using
    $\displaystyle
    x = 2sin\theta\
    $
    Last edited by nath_quam; Jul 3rd 2006 at 03:09 AM.
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  6. #6
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    Hello, Nath!

    $\displaystyle \int^1_{-1}\!\sqrt{4-x^2}\,dx$ . using $\displaystyle
    x = 2\sin\theta$

    We can do it, but it takes a lot of work
    and you're expected to be very familiar with Trigonometry.

    I'll go through the whole thing in baby-steps . . .

    Let $\displaystyle \x = 2\sin\theta\quad\Rightarrow\quadd x = 2\cos\theta\,d\theta$

    Note that: .$\displaystyle \sqrt{4 - x^2} \:= \:\sqrt{4 - 2\sin^2\theta} \:=$ $\displaystyle \sqrt{4(1 - \sin^2\theta)} \:=\:\sqrt{4\cos^2\theta} \:=\:2\cos\theta$

    Substitute: .$\displaystyle \int(2\cos\theta)(2\cos\theta\,d\theta) \:=\:4\int\cos^2\theta\,d\theta$

    Double-angle identity: .$\displaystyle 4\int\frac{1 + \cos2\theta}{2}\,d\theta \;= \;2\int(1 + \cos2\theta)\,d\theta$

    Integrate: .$\displaystyle 2\left[\theta + \frac{1}{2}\sin2\theta\right] $

    Half-angle identity: .$\displaystyle 2\left[\theta + \frac{1}{2}(2\sin\theta\cos\theta)\right] \:= \:2\left[\theta + \sin\theta\cos\theta\right] $


    Now we must back-substitute.
    We have: .$\displaystyle x = 2\sin\theta\quad\Rightarrow\quad\sin\theta = \frac{x}{2}\quad\Rightarrow\quad\theta = \arcsin\frac{x}{2}$
    . . and: .$\displaystyle \cos\theta = \frac{\sqrt{4-x^2}}{2}$


    So we have: .$\displaystyle 2\left[\,\arcsin\frac{x}{2} + \frac{x\sqrt{4-x^2}}{4}\,\right]^1_{\text{-}1} $

    $\displaystyle 2\left[\arcsin\left(\frac{1}{2}\right) + \frac{1\cdot\sqrt{4-1^2}}{4}\right] -$$\displaystyle 2\left[\arcsin\left(-\frac{1}{2}\right) + \frac{(-1)\sqrt{4 - (-1)^2}}{4}\right] $

    . . $\displaystyle = \;2\left(\frac{\pi}{6} + \frac{\sqrt{3}}{4}\right) - 2\left(-\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)$

    . . $\displaystyle = \;\frac{\pi}{3} + \frac{\sqrt{3}}{2} + \frac{\pi}{3} + \frac{\sqrt{3}}{2}$

    . . $\displaystyle = \;\frac{2\pi}{3} + \sqrt{3} \;\approx\;3.826446$

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  7. #7
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    Quote Originally Posted by Soroban

    Integrate: .$\displaystyle 2\left[\theta + \frac{1}{2}\sin2\theta\right] $
    At this stage couldn't you x sub 1 and -1 into
    $\displaystyle
    x = 2\sin\theta\
    $
    and get the answer that way
    so you end up with
    $\displaystyle
    2\left[\theta + \frac{1}{2}\sin2\theta\right]^\frac{\pi}{2}_{\frac{3\pi}{2}}
    $
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  8. #8
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    hello, Nath!

    Your idea is a good one . . .


    At this stage couldn't you x sub 1 and -1 into $\displaystyle x = 2\sin\theta$ and get the answer that way . . . Yes!

    so you end up with: $\displaystyle 2\left[\theta + \frac{1}{2}\sin2\theta\right]^\frac{\pi}{2}_{\frac{3\pi}{2}}$ . . . but your limits are incorrect.

    We have: .$\displaystyle 2\sin\theta = x$

    For $\displaystyle x = -1:\;\;2\sin\theta\,=\,-1\quad\Rightarrow\quad\sin\theta\,=\,-\frac{1}{2}\quad\Rightarrow\quad\theta = -\frac{\pi}{6}$

    For $\displaystyle x = 1:\;\;2\sin\theta\,=\,1\quad\Rightarrow\quad\sin \theta\,=\,\frac{1}{2}\quad\Rightarrow\quad \theta = \frac{\pi}{6}$


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