Tangent Line and area of a triangle

Question

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.

Answer 1

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?

Answer 2

The function $y$ you have is a line. The first step after that is to calculate intersections with the axes (note the sign): $$y=e^a+e^a(x-a)$$ The intersection with the vertical axis is at $x=0$, so the corresponding $y$ is $y_0=e^a(1-a)$. Similarly, the intersection with the horizontal axis is at $y=0$, so $$e^a+e^a(x_0-a)=0$$ or $x_0=a-1$. Note that $a\le0$. Then the area of the triangle is $A=\frac 12 x_0y_0$. You write this in terms of $a$, and you minimize the expression.