Setting $x=0,$ we obtain
\[0 = f(0) = (1 - 0) f'(0),\]which means $f(0) = 0.$
[asy]
unitsize(1 cm);
real func (real x) {
return(exp(-x^2));
}
pair T(real x) {
real dy = 2*x*exp(-x^2);
real dx = 1;
real m = dy/dx;
return((x - m,x + m));
}
draw(graph(func,0.03,1.3));
draw((0,0)--(0,1),dashed);
draw(T(0));
draw(T(0.1));
draw(T(0.2));
draw(T(0.3));
draw(T(0.4));
draw(T(0.5));
draw(T(0.6));
draw(T(0.7));
draw(T(0.8));
draw(T(0.9));
draw(T(1));
draw(T(1.1));
draw(T(1.2));
draw((0,0)--(1.7,0),dashed);
label("$A$", (0,0), W);
label("$B$", (0,func(0)), NW);
label("$T$", (0.8,0), S, UnFill);
[/asy]
Let $T$ be the point of tangency. Then by the Two Chord Theorem, $AB$ pass through a fixed point $T.$ The slope of $AB$ is equal to the slope of the tangent line at $T.$
Hence, all the tangent lines are concurrent, so the graph of $f(x)$ must approach a horizontal line $L$ as $x \to \infty.$
Let $k = f'(x).$ Let $T_x$ be the point on the graph where the tangent line meets the $x$-axis, and let $B_x = (0,f(x))$ be the point where the tangent line meets the $y$-axis. Then by the Two Chord Theorem, all the lines $T_x B_x$ pass through a point $T$ on the graph.
Let $y = f(x)$ and $x = g(y).$ Then the slope of $\overline{T_x B_x}$ is
\[\frac{f(x) - 0}{0 - g(f(x))} = -\frac{g(y)}{g'(y)} = -\frac{x}{f'(x)},\]so
\[\frac{dy}{dx} = -\frac{x}{f'(x)}.\]By the Product Rule,
\begin{align*}
-f'(x) + xf''(x) &= 0, \\
f''(x) &= \frac{f'(x)}{x}.
\end{align*}Let $u = f'(x).$ Then $u = \frac{f(x)}{x},$ so
\[u' = \frac{xf'(x) - f(x)}{x^2}.\]Then using the fact that $f'(x) < 0,$ $f(x) > 0,$ and $f(0) = f'(0) = 0,$ we find
\begin{align*}
u' &= \frac{xf'(x) - f(x)}{x^2} \\
&= \frac{xf'(x) - x f(x)/x}{x^2} \\
&= \frac{x f'(x) - x f'(x)/x}{x^2} \\
&= 0.
\end{align*}Therefore, $u = C$ for some constant $C.$ Hence,
\[f'(x) = Cx.\]If $C = 0,$ then $f'(x) = 0$ for all $x,$ which means $f(x)$ is constant. But then $f(x)$ is not positive.
Otherwise, let $C \neq 0.$ Then we can write
\[f'(x) = Cx\]as $\frac{f'(x)}{f(x)} = Cx.$ Integrating both sides, we get
\[\ln |f(x)| = \frac{1}{2} Cx^2 + D\]for some constant $D.$ Then
\[|f(x)| = e^D e^{\frac{1}{2} Cx^2}.\]Since $f(x) > 0,$ $e^D$ is positive. We can absorb $e^D$ into the constant $E,$ so
\[f(x) = E e^{\frac{1}{2} Cx^2}.\]Since $f(0) = 0,$ $E = 0,$ so $\boxed{f(x) = 0}$ is the only solution.
The function $f(x),$ defined for $x > 0,$ is positive, differentiable, and decreasing. The function $f(x)$ has the following property: Let $P$ be a point on the graph, and draw the tangent to the graph at $P.$ Let the tangent meet the $x$-axis and $y$-axis at $A$ and $B,$ respectively. Then the length of chord $AB$ is constant.
Find all functions $f(x)$ that have this property.
1 answer