Let $\displaystyle f(x)=\sqrt{x}$, $\displaystyle x>0$.

Then, $\displaystyle f^{\prime}(x) = \frac{1}{2\sqrt{x}}$ , $\displaystyle x>0$.

Since $\displaystyle f$ is continuous on $\displaystyle [3,4]$ and differentiable on $\displaystyle (3,4)$, there exists $\displaystyle c\in(3,4)$ such that

$\displaystyle f^{\prime}(c)=\frac{f(4)-f(3)}{4-3}=f(4)-f(3)$.

So, there exists $\displaystyle c\in(3,4)$ such that

$\displaystyle \sqrt{4}-\sqrt{3}=2-\sqrt{3}=\frac{1}{2\sqrt{c}}$.

Now we prove the boundaries:

$\displaystyle 3<c<4 \implies \sqrt{3}<\sqrt{c}<2 \implies \frac{1}{2}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{3}} \implies \frac{1}{4}<\frac{1}{2\sqrt{c}}<\frac{\sqrt{3}}{6} $

Since $\displaystyle \frac{1}{2\sqrt{c}}=2-\sqrt{3}$ and the problem gives us that $\displaystyle \sqrt{3}<1.8$ , the last inequality implies

$\displaystyle \tiny{\frac{1}{4}<2-\sqrt{3}<\frac{1.8}{6} \implies 0.25-2<-\sqrt{3}<0.3-2 \implies -1.75<-\sqrt{3}<-1.7 \implies 1.7<\sqrt{3}<1.75}$.