Bolzano theorem and interval solution I have a function $f$ witch is continuous at $[1.4]$. I also have that $f(x)\neq 0,\forall x\in [1,4]$ and $f(1)\cdot f(2)\cdot f(4)=8$. I have already proved that $f(x)>0,\forall x\in [1.4]$.
Now I want:


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*To prove that the equation $f(x)=2$ has at least one solution at $[1,4]$. So I thought to use Bolzano's theorem. Let function $g$ with type $g(x)=f(x)-2$ which is continuous at $[1,4]$, $g(1)=f(1)-2$ and $g(4)=f(4)-2$. Now I need to prove that $g(1)\cdot g(4)\le 0$, isn't it right? But if it is, how can I prove that?

*To prove that the equation $f(x)=x$ has at least one solution at $[1,4]$. In here I use the same method as before?


I know one standard way to state the intermediate value theorem is that if $f(a)<0$ and $f(b)>0$ then $f(x)=0$ for at least one $x$ in $(a, b)$, but in this case I think you're obscuring the problem by reformulating it that way. The point is that if $f(x)$ is never equal to $2$ on $[1, 4]$, it's either identically less than $2$ or identically greater than $2$. Why are neither of those options possible?
For 2, if $f(x)$ is never equal to $x$, then we either have $f(x)>x$ on $[1, 4]$ or $f(x)<x$ on $[1, 4]$. Why does that contradict the given condition on $f$?
