Show that if $f : [1, 2] →\Bbb R$ is a continuous then there exists $\gamma\in (1, 2)$ with $f(\gamma) = \frac{1}{1 − \gamma}+ \frac{1}{2 − \gamma}.$ Show that if $f : [1, 2] →\Bbb R$ is a continuous function then there exists $\gamma\in (1, 2)$ such that $$f(\gamma) = \frac{1}{1 − \gamma}+ \frac{1}{2 − \gamma}.$$
I have been trying to understand what the question wants me to do but I can't figure it out. Please help!

Consider the function $$g:x\mapsto  f(x)-\frac{1}{1 − x}- \frac{1}{2 − x},1<x<2.$$ Since $\lim_{x\to 1+}f(x)=f(1)$ and $\lim_{x\to 2-}f(x)=f(2)$ we have, $$\lim_{x\to 1+}g(x)=+\infty,$$$$\lim_{x\to 2-}g(x)=-\infty.$$ So by intermediate value theorem we have $\gamma\in (1,2)$ such that, $g(\gamma)=0$.
