= Solution
For $f(x)=x^{1/4}$, the first two <derivatives> are $f'(x)=\tfrac14x^{-3/4}$ and $f''(x)=-\tfrac3{16}x^{-7/4}$. At $16$ they give
$$
f(16)=2,\qquad f'(16)=\frac1{32},\qquad f''(16)=-\frac3{2048}.
$$
Use the first three <Taylor expansion> terms with increment $17-16=1$:
$$
\boxed{17^{1/4}\approx2+\frac1{32}-\frac3{4096}=2.030517578125.}
$$
The positive third <derivative> shows that this quadratic approximation underestimates the true value. The <Taylor remainder> is at most $7/262144\approx2.67\times10^{-5}$, since $f^{(3)}(x)=(21/64)x^{-11/4}$ decreases on $[16,17]$ and the cubic remainder is $f^{(3)}(\xi)/6$.
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