= Solution
For $f(x)=x^{1/5}$, the first two <derivatives> are $f'(x)=x^{-4/5}/5$ and $f''(x)=-4x^{-9/5}/25$. Since $32=2^5$,
$$
f(32)=2,\qquad f'(32)=\frac1{80},\qquad f''(32)=-\frac1{3200}.
$$
The first three terms of the <Taylor expansion>, with $x-32=1$, therefore give
$$
\boxed{33^{1/5}\approx2+\frac1{80}-\frac1{6400}=\frac{12879}{6400}=2.01234375.}
$$
For a check on accuracy, $f^{(3)}(x)=36x^{-14/5}/125$ is positive and decreasing on $[32,33]$. The <Taylor remainder> is consequently positive and less than $f^{(3)}(32)/6=3/1024000<3\times10^{-6}$. The approximation is a slight underestimate.
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