If f: [2,5] $\rightarrow [4,13]$ is a continuous function prove that there is $c\in [2,5]$ such that $f(c) = 3c-2$

Question

If $f: [2,5] \rightarrow [4,13]$ is a continuous function prove that there is $c\in [2,5]$ such that $f(c)= 3c-2$.

I understand that the extreme value theorem is important in this question and that at most if $c=5$ then $3*5-2=13$, but I am struggling to explain this in a clear and orderly manner. Any advice on how to do Extreme Value Theorem proofs?

Answer 1

Well: $$g(x)=3x-2\implies g(2)=4, g(5)=13$$

and, given $f:[2,5]\to [4,13]$, this means $4\leq f(t)\leq 13$, i.e. $g(2)\leq f(t)\leq g(5)$ for every $t\in[2,5]$

We can infer that, specifically, $f(2)\geq g(2)$ and $f(5)\leq g(5)$, and the only way this is satisfied for a continuous function $f$ is if it has an intersection with $g$ in the range $[2,5]$

The below picture demonstrates. The space $([2,5]\times [4,13])$ on which the function is defined is the green rectangle, and $g$ is the blue line, while $f$ is any continuous curve (with no breaks or jumps) that goes from the left side of the rectangle to the right side without leaving it. It's easy to see it's impossible for this curve to not cross or touch the blue line on the way.

Answer 2

In case $f(2)=4+p$ and $f(5)=13-q, p,q>0$, let $g(x)=f(x)-3x+2 \implies g(2)=4+p-4=p, g(5)=13-q-13=-q$, then $g(2)g(5)=-pq<0$. So by IVT it follows that $g(c)=0 \implies f(c)=3c-2$ for some $c\in (2,5)$.

If $pq=0$ the statement satisfied trivially.