 Solution

Expand $f(s)=\sqrt{s}$ about $s=25$, with increment one. Since $f^{\prime }(25)=1/10$ and $f^{\prime \prime }(25)=-1/500$, the second-order Taylor series approximation isThe third derivative is $3/(8s^{5/2})>0$, decreasing on $[25,26]$. The Taylor remainder is therefore positive and less than $f^{\prime \prime \prime }(25)/6=1/50000$, so $0<\sqrt{26}-5.099<0.00002$.