 Solution

For $f(x)=x^{1/4}$, the first two derivatives are $f^{\prime }(x)=\frac{1}{4}{x}^{-3/4}$ and $f^{\prime \prime }(x)=-\frac{3}{16}{x}^{-7/4}$. At $16$ they giveUse the first three Taylor expansion terms with increment $17-16=1$:The positive third derivative shows that this quadratic approximation underestimates the true value. The Taylor remainder is at most $7/262144\approx 2.67\times 10^{-5}$, since $f^{(3)}(x)=(21/64)x^{-11/4}$ decreases on $[16,17]$ and the cubic remainder is $f^{(3)}(\xi )/6$.