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? 

Let $h(x) = f(x) - 4x + 4$. $h$ is continuous on $[1,3]$. Moreover:
$h(1) = f(1) \ge 0$ and $h(3) = f(3) - 8 \le 0$ since $\operatorname{Range}(f) = [0,8]$. 
Thus, by IVT..
