If $f$ is an even function defined on $(-5,5)$ on the interval, then find four real values of x satisfying $f(x)=f(\frac{x+1}{x+2})$.
My book gives the answer as $\frac{\pm3 \pm\sqrt{5}}{2}$
And the solution is given as:
Since, f is an even function,
then, $f(x)=f(-x)$,$\forall x\in (-5,5)$
$$\implies f(x)=f(\frac{1-x}{2-x}) \implies x=\frac{1-x}{2-x} \implies x=\frac{3\pm\sqrt{5}}{2}$$
Again, $$f(-x)=f(\frac{x+1}{x+2}) \implies -x=\frac{x+1}{x+2} \implies x=\frac{-3\pm\sqrt{5}}{2}$$
What I can't understand is that how can we say $\implies f(x)=f(\frac{1-x}{2-x}) \implies x=\frac{1-x}{2-x}$? It is nowhere said that $f(x)$ is a one-one function then how can we say so? Also if we do use this logic then $f(x)=f(\frac{x+1}{x+2})$
which gives $x=\frac{-1\pm\sqrt{5}}{2}$ which is not given as a solution.
Best Answer
You are correct; we can't conclude that $x=\frac{1-x}{2-x}$ just from $f(x) = f\left(\frac{1-x}{2-x}\right)$. But we don't need to; we need to make the opposite inference: If $x = \frac{1-x}{2-x}$, then $f(x) = f\left(\frac{1-x}{2-x}\right)$, which is what we are looking for.
There might of course be many other $x$ for which $f(x) = f\left(\frac{1-x}{2-x}\right)$ but $x\ne\frac{1-x}{2-x}$. For example, $f$ might be a constant function so that $f(x) = f\left(\frac{1-x}{2-x}\right)$ for all $x\ne 2$. But the question only asks us to find four such values, and this we can certainly do.