Use a linear approximation to estimate the number $64.07^{2/3}$ That's all they give you. I tried putting it into the linear approximation equation of:
$$
f(a)+f'(a)(x-a)
$$
but I get almost the same value as $64.07^{2/3}$, which is around $16.0116$. Just not sure how else to go about doing this problem. Thanks.
 A: Consider function $f(x)=x^{2/3}$ and establish, from definitions, Taylor series at $x=64$ since $64^{2/3}=16$.
You will get for first order (linear approximation)  $$f(x)=16+\frac{x-64}{6}+O\left((x-64)^2\right)$$ from which $x=\frac{9607}{600}\approx 16.01166667 $. This is exactly what you did.
For more accuracy, you could use the second order expansion (quadratic approximation) would be   $$f(x)=16+\frac{x-64}{6}-\frac{(x-64)^2}{2304}+O\left((x-64)^3\right)$$ from which $x=\frac{368908751}{23040000}\approx 16.01166454$
For ten significant digits, the exact solution is $\approx 16.01166454$.
A: An equivalent formula
for $x^a$
is
$(1+x)^a
\approx 1+ax
$ 
for small $x$.
Therefore,
for small $v$,
$(u^{a}+v)^{1/a}
=u(1+\frac{v}{u^a})^{1/a}
\approx u(1+\frac{v}{au^a})
= u+\frac{uv}{au^a}
= u+\frac{u^{1-a}v}{a}
$.
In this case,
$a = \frac32$,
so
$(u^{3/2}+v)^{2/3}
\approx u+\frac{u^{-1/2}v}{\frac32}
=u+\frac{2v}{3\sqrt{u}}
$.
If $u=16$
and $v = .07$,
this gives
$16+\frac{.14}{12}
=16.01166...
$.
