I have $f(x) = e^{-x}$. There is a tangent to the function, the tangent point is in the first quadrant. Find the area between the axis $y$, $x$ and the tangent where the area is maximum.
So obviously the area is a triangle. However I don't have idea in how to begin. I thought about putting the triangle's vertexes as $(0,y_1)$ and $(x_1,0)$.
Consider the tangent at point $\displaystyle (x_0, e^{-x_0})$ to the curve. The slope is $\displaystyle -e^{-x_0}$.
So the equation of the tangent may be written: $\displaystyle y - e^{-x_0} = -e^{-x_0}(x-x_0)$
The $\displaystyle y$ and $\displaystyle x$ intercepts of the tangent line are $\displaystyle (0,e^{-x_0}(x_0 + 1))$ and $\displaystyle (x_0 + 1, 0)$ respectively giving the area of the bounded triangle as $\displaystyle A(x_0) = \frac 12e^{-x_0}(x_0 + 1)^2$
Set $\displaystyle \frac{dA}{dx_0} = 0$ to get $\displaystyle x_0 = \pm 1$. We ignore the negative value as we're told it's in the first quadrant. So $\displaystyle x_0 = 1$.
Prove this stationary point is a maximum by showing $\displaystyle \frac{d^2A}{dx_0^2}\biggr{|}_{x_0=1}<0 $
So finally, the required area is $\displaystyle \frac 12 \cdot e^{-1}\cdot 4 = \frac 2e$