Tangentially touching functions {y=ln(f(x)) and y=f(x)/e}

Question

I found some really interesting graphs like these:

So basically through these graphs I want to ask that if we take 2 functions in the format of $$f_1 = ln(f(x))$$ and $$f_2 = \frac{f(x)}{e}$$ do they always intersect tangentially?

Answer 1

Well, for two functions to touch tangently in $x_0$, 2 conditions must meet (or 3 if you want to avoid intersection, but I'm not going to include this case):

1. $$f_1(x_0) = f_2(x_0)$$
2. $$f_1'(x_0) = f_2'(x_0)$$

If you substitute your functions in question to the conditions, we get: $$\ln(f(x_0)) = f(x_0) / e$$ $$\frac{f'(x_0)}{f(x_0)} = f'(x_0) / e \implies f(x_0) = e ~\text{or}~f'(x_0)=0$$

Now if there exists $x_o\in\mathbb{R}$ s.t. $f(x_0) = e$ it follows that the first equality also holds,so you have a touching point, but not for a general $f$. For example $f(x)=1$

Edit: For the third condition mentioned above check out @Martin R's answer. He proves that $f_1(x) \le f_2(x) (\forall x \in \mathbb R)$. So the curves not only share the same tangent line at points where $f(x)=e$ but they are also "just touching" and not intersecting.

Answer 2

The inequality $\ln (u) \ge u/e$ holds for all $u > 0$, because the logarithm is concave, so that its graph lies below the tangent at $u=e$: $$ \ln (u) \le \ln (e) + (u-e)\frac 1e = \frac ue \, , $$ with equality if and only if $u=e$.

Therefore, if $f(x) > 0$ then $$ h(x) = f_1(x) - f_2(x) = \ln f(x) -\frac{f(x)}{e} \le 0 \, , $$ so that the graph of $f_1$ is never above the graph of $f_2$.

If the graphs “meet” at a point $x^*$ then $h$ has a maximum at that point. If $f$ is differentiable at $x^*$ then $h'(x^*) = 0$ follows, i.e. $f_1'(x^*) = f_2'(x^*)$.

That explains why $f_1$ and $f_2$ “meet tangentially”: they have the same slope at the intersection points.