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- May 19th 2008, 11:09 AMnnnnn[SOLVED] Inverse Functions
delete now plz

- May 19th 2008, 11:32 AMTheEmptySet

Let a be the point of the tangent line then

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

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

So the slope at point a is

$\displaystyle m=-e^{-a}$

Now we can find the equation of the line through the point $\displaystyle (a,e^{-a})$

$\displaystyle 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

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

$\displaystyle ((a+1),0)$

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

So its area is

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

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

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

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

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

Yeah!!!(Rock) - May 19th 2008, 01:35 PMnnnnn
sweet thnx