The line tangent to the graph $f(x)=e^x$ at a point $x \le 0$ intersects both axes forming a triangle. Find $x \le 0$ that minimizes the area of this triangle and the value of the corresponding area.
In this exercise I know that the tangent is of the form
$$y=f(a)-f'(a)\cdot(x-a)$$
But after doing operations, I realize that the number $x$ can be as small as I want, that this area will continue to be smaller.
Best Answer
The equation of the tangent line is given by $y-a=f'(b)(x-b)=e^b(x-b),$ where $x\le 0$ and $f(x)=e^x,$ and also the point of tangency is given by $(b,a).$ Thus the intercepts on both axes, taken absolutely, give the lengths of the legs of the right triangle, namely $|a-be^b|$ and $|-ae^{-b}+b|$ respectively.
Hence the area of this triangle is given by $$\frac12|(a-be^b)(b-ae^{-b})|,$$ where $a$ is given in terms of the $x$-coordinate $b$ by $a=e^b.$ Thus our objective function in terms of $b$ alone is given by $$\frac{e^b|(1-b)(b-1)|}{2}=\frac{e^b(b-1)^2}{2},$$ defined for all $b\le 0.$
Can you now see that this function never vanishes on its domain? Also, can you now find the minimum of this function on its domain?