A question about bijective functions Suppose that $f: [1,3]$$\rightarrow$$[0,8]$ is continuous. Show that there is some $x$$\in$$[1,3]$ such that $f(x)+4=4x$.
I need some advice on how to get started. Do i need to use IVT for this problem? Do I need to show that f is increasing while x is increasing? 

EDIT: Let $h(x)= \frac {f(x)+4}{4}: [0,8] \rightarrow [0,8] $ with $f:[1,3] \rightarrow [0,8] $ as given.
Then $h(x)$ is a continuous map from a compact convex set to itself, and so it has a fixed point *, so that there is $x$ with $h(x)= \frac {f(x)+4}{4}=x  $ , so that $$ f(x)+4 =4x  $$ has a solution in $[0,8]$
*:https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem . Note that we only need the map to be into and not onto $[0,8]$. Note that the map is into but  not onto.
