Approximating Functions with Polynomials in Taylor Series I'm having difficulty with a series of problems I've been working on and I can't seem to find a straightforward explanation of how to solve them. The problem is approximating functions.
$f(x)=x^{1/3}$, $a=8$, approximate $7.5^{1/3}$
I was able to calculate $p_1(x)=2+(1/12)(x-8)$, $p_2(x)=2+(1/12)(x-8)-(1/288)(x-8)^2$
I know need to use the linear approximating polynomial and the quadratic approximating polynomial to come up with an answer. I'm unsure of how to use the $7.5^{1/3}$ to come up with an answer. Any help would be greatly appreciated. 
 A: Hint : Just plug $x=7.5$ into the approximation-polynomials.
The exact result and the approximations are (calculation with PARI/GP) :
? 7.5^(1/3)
%4 = 1.957433820584431797712468030
? 2+1/12*(7.5-8)
%5 = 1.958333333333333333333333333
? 2+1/12*(7.5-8)-1/288*(7.5-8)^2
%6 = 1.957465277777777777777777778
?

A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
7.5^{1/3} & = \pars{15 \over 2}^{1/3} = \pars{60 \over 8}^{1/3} =
\pars{64 - 4 \over 8}^{1/3} =
\bracks{{64 \over 8}\pars{1 - {1 \over 16}}}^{1/3} =
2\pars{1 - {1 \over 16}}^{1/3}
\\[3mm] & \approx 2 - {1 \over 24} - {1 \over 1152} - {5 \over 165888} - {5 \over 3981312}
\approx 1.95743\color{#f00}{388109246}
\end{align}

Relative Error $\ds{\sim 10^{-6}\ \%}$.

