Let $f:(0,+\infty)\to (0,+\infty)$ be a differentiable function such that $f'(x)\not=0$ with the following property: the area of the triangle formed by the tangent line of $C_f$ at a point $M(x_0,f(x_0))$, $x'x$ and $y'y$ is constant and independent of $x_0\in (0,+\infty)$. Find all functions $f$ with the above properties.
I know that $f(x)=\frac{1}{x}, x>0$ satisfies the above conditions.
Attempt
Let $\varepsilon_{x_0}:\ y-f(x_0)=f'(x_0)(x-x_0)$ be the equation of a tangent line of $C_f$ at a point $M(x_0,f(x_0))$ and $A\bigg(\frac{x_0f'(x_0)-f(x_0)}{f'(x_0)},0\bigg)$, $B(0,-x_0f'(x_0)+f(x_0))$
the intersection points of $\varepsilon_{x_0}$ with the axes. Then $(OAB)=c=\text{constant}\Leftrightarrow (x_0f'(x_0)-f(x_0))^2=-2cf'(x_0)$.
Questions
- Is there any reference of the above problem in the literature?
- Is there an easy way to solve the differential equation $(xf'(x)-f(x))^2=-2cf'(x)$?
Best Answer
Just wanted to connect and expand a little bit my comments.
Differentiating the equation $$\left[xf'(x) - f(x)\right]^2=-2cf'(x)\tag{1}\label{1}$$ you get $$f''(x)\left[x^2f'(x)-xf(x)+c\right]=0.$$ So either $$f''(x) = 0,$$ yielding $$f(x) = k_1 x + k_2,$$ which however does not satisfy the requirement $ f: \ (0,+\infty)\to (0,+\infty)$, or $$f'(x) = \frac1{x} f(x) -\frac{c}{x^2},$$ a linear ODE whose solutions are $$f(x) = kx + \frac{c}{2x}.$$ Plugging in these solutions into \eqref{1} we get $k=0$. So the only twice differentiable functions that satisfy the hypotheses are of the form $$\boxed{f(x) = \frac{c}{2x}},\tag{2}\label{2}$$ where $c>0$ is the triangle constant area. I wonder what happens if you remove the requirement on twice-differentiability.
EDIT
Following the comment by Alex Ravsky one can note that equation \eqref{1} can be written in the form $$x^2[f'(x)]^2-2(xf(x)-c)f'(x)+f^2(x) = 0,$$ which implies that at point $x>0$ the derivative can be computed as $$f'(x) = \frac{f(x)}{x}-\frac{c}{x^2} \pm \frac{\sqrt{c[c-2xf(x)]}}{x^2}.$$ Observe that this forces the condition $f(x)\leq \frac{c}{2x}$, for all $x>0$. Using now the derivative of composite (differentiable) functions we can determine that the second derivative of $f(x)$ does exist and is equal to $$f''(x) = \frac{xf'(x)-f(x)}{x^2}+\frac{3c}{x^3}\pm\frac{-x\ \boxed{\frac{\left\{c[c-2xf(x)]\right\}'}{\sqrt{c[c-2xf(x)]}}}-2\sqrt{c[c-2xf(x)]}}{x^3},\tag{3}\label{3}$$ valid for all points $x$ such that $f(x)<\frac{c}{2x}$, unless $xf(x)$ is constant (in which case the third summand in \eqref{3} is just $\pm k/x^3$, for some constant $k$). The latter case confirms the solution \eqref{2}. The former does not apply, since we already know that the only function satisfying the original requirement and that has second derivative is $f(x) = \frac{c}{2x}$.
In conclusion, no other function $(0,+\infty)\to (0,+\infty)$, aside from \eqref{2}, guarantees that the tangent at any point forms with the axes a triangle with constant area.