Suppose there is a function $f(x)=\frac{x^2-2x+b}{x^2+2x+b}$ (the problem doesn't specify, but I am assuming $b$ is a real) that has a minimum value of $\frac{1}{2}$. What is the maximum value of $f(x)$?
My first instinct was to divide out everything, getting that $f(x)=1-\frac{4x}{x^2+2x+b}$. From there, I'm not sure what to do.
I am looking for a solution that does not involve calculus.
Consider $$y=\frac{x^2-2x+b}{x^2+2x+b}$$ Or $$(y-1)x^2+2(y+1)x+b(y-1)=0$$ In this quadratic equation in $x$, for those values of $y$ which exist in range, there should exist real value(s) of $x$.
Hence, Discriminant $\geq 0 $. $$4(y+1)^2-4b(y-1)^2\geq0$$ $$(1-b)y^2+2y(1+b)+(1-b)\geq0$$
Now this is a quadratic in $y$ which will provide us with the values of $y$ in range. More specifically, those values of $y$ satisfying this inequality will be in the range.
We know that $y=\frac{1}{2}$ is the starting point of the range, so the inequality must be exactly $0$ at $y=\frac{1}{2}$. $$\frac{1-b}{4}+(1+b)+(1-b)=0$$ That is $b=9$.
That provides us the other extreme of the inequality by solving it, $$-8y^2+20y-8\geq0$$ $$2y^2-5y+2\geq0$$ So $$\frac{1}{2}\leq y \leq 2$$
