Suppose $f: [1,2] \to [5,7]$ is continuous. Show that $f(c)=2c+3$ for some $c \in [1,2]$.
First note $f(1)=5$ and $f(2)=7$. By the IVT, all values $c \in [1,2]$ are hit. I'm just wondering how to put all of these facts together to arrive at $f(c)=2c+3$ for all $c \in [1,2]$.