1. ## [SOLVED] Inverse Functions

delete now plz

2. Originally Posted by nnnnn
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!!!

3. sweet thnx