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Math Help - Functions

  1. #1
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    Functions

    A function
    <br />
y = f(x)<br />
    under what geometrical conditions would
    <br />
f(x) = f^{-1} (x) ?<br />
    and whats an example of this function for which
    <br />
f(x) = f^{-1} (x) ?<br />

    &

    does
    <br />
\int^{1}_{-1} \sqrt\(4-x^{2}\) <br />

    <br />
= 2{\pi}<br />
    Last edited by nath_quam; July 2nd 2006 at 05:37 PM.
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  2. #2
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    Quote Originally Posted by nath_quam
    A function
    <br />
y = f(x)<br />
    under what geometrical conditions would
    <br />
f(x) = f^{-1} (x) ?<br />
    For example, the identity function,
    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
    <br />
\int^{1}_{-1} \sqrt\(4-x^{2}\) <br />

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

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

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

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

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

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  5. #5
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    Would someone be able to solve this

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

    \int^1_{-1}\!\sqrt{4-x^2}\,dx . using <br />
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 \x = 2\sin\theta\quad\Rightarrow\quadd x = 2\cos\theta\,d\theta

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

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

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

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

    Half-angle identity: . 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: . x = 2\sin\theta\quad\Rightarrow\quad\sin\theta = \frac{x}{2}\quad\Rightarrow\quad\theta = \arcsin\frac{x}{2}
    . . and: . \cos\theta = \frac{\sqrt{4-x^2}}{2}


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

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

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

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

    . . = \;\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: . 2\left[\theta + \frac{1}{2}\sin2\theta\right]
    At this stage couldn't you x sub 1 and -1 into
    <br />
x = 2\sin\theta\<br />
    and get the answer that way
    so you end up with
    <br />
2\left[\theta + \frac{1}{2}\sin2\theta\right]^\frac{\pi}{2}_{\frac{3\pi}{2}}<br />
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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 x = 2\sin\theta and get the answer that way . . . Yes!

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

    We have: . 2\sin\theta = x

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

    For 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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