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Math Help - [SOLVED] Inverse Functions

  1. #1
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    [SOLVED] Inverse Functions

    delete now plz
    Last edited by nnnnn; May 19th 2008 at 01:36 PM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by nnnnn View Post
    What is the area if the largest triangle in the first quadrant with 2 sides on the axes and third side tangent to the curve y=e^-x

    Let a be the point of the tangent line then

    (a,e^{-a}) is a point on the graph

    y=e^{-x} \to y'=-e^{-x}

    So the slope at point a is

    m=-e^{-a}
    Now we can find the equation of the line through the point (a,e^{-a})

    y-e^{-a}=-e^{-a}(x-a) \iff y=-e^{-a}x+(a)e^{-a}+e^{-a}

    We can now find the x and y intercepts of the equation of the line

    (0,e^{-a}(a+1)) and

    ((a+1),0)

    These are the lengths of the base and height of the triangle

    So its area is

    A(a)=\frac{1}{2}(a+1)(e^{-a}(a+1))

    A(a)=\frac{1}{2}(a+1)^2e^{-a}

    \frac{dA}{da}=(a+1)e^{-a}-\frac{1}{2}(a+1)^2e^{-a}=e^{-a}(a+1)(1-\frac{1}{2}a-\frac{1}{2})

    setting each factor equal to zero and solving gives a=1 or a=-1

    since we want the part in the first quadrant we don't use a=-1

    A(1)=\frac{1}{2}(1+1)(e^{-1}(1+1))=2e^{-1}=\frac{2}{e}

    Yeah!!!
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    sweet thnx
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