How do you find the fixed points of a rational function of f(x)=40x+145/x+32.5?

Question

not sure whether to solve for a quadratic or what exactly? any help would be much appreciated.

Answer 1

f(x)=(40x+145)/(x+32.5).

Critical points for x are -145/40 (when f(x) is zero) and -32.5 (when f(x) is undefined, so x=-32.5 is the vertical asymptote). The horizontal asymptote is 40x/x=40, that is, y=40.

The fixed points of f(x) are when f(x)=x, so

x=(40x+145)/(x+32.5).

x²+32.5x=40x+145,

x²-7.5x-145=0,

2x²-15x-290=0.

x=(15±√(225+2320))/4=(15±50.448)/4=3.75±12.612=16.362,-8.862 approx.