I was dabbling in some recreational math and I stumbled upon this problem:
Say there are two arbitrary functions $f(x)$ and $g(x)$. Create the normal line to $g(x)$ at $x_g$; now find an expression for $x_f$, the x-value of where the normal line hits $f(x)$, in terms of $x_g$. So essentially, find $x_f(x_g)$.
The equation of the normal line is simple to find: $y=\frac{-1}{g'(x_g)}(x-x_g)+g(x_g)$
My main method for trying to solve this problem has been with geometric constructions involving the angles shown and lengths of the triangles shown. The zero of the normal line is $x=g(x_g)g'(x_g)+x_g$ and the zero of the tangent line is $x=x_g-\frac{g(x_g)}{g'(x_g)}$. Given these lengths and others in the diagram, I have not been able to solve the problem by finding some proportion or geometric expression relating $x_f$ to $x_g$.
My question is if such an expression is even possible to create and if so, how would it be done? I apologize for not having more information on the problem and how it could be solved, this is all I've gotten after hours of working this problem. Thank you in advance for the help.
You claim:
Let the right-hand side of your equation be $h(x)$. Then, assuming it is correct, you can then find the abscissa of the point of intersection of the line given by $y=h(x)$ with the locus of points $(x,y)$ determined by $y=f(x)$ by solving the following system of equations for $x$: \begin{cases} y=h(x);\\ y=f(x). \end{cases} Please comment if you want me to elaborate further.