Quadratic approximation of $9^{1/3}$ Find a quadratic approximation of the cube root of $9$ by using the equation $9=8(1+\frac 18)$, and estimate the difference between the exact value and the approximation.
How am I supposed to find the quadratic approximation of this formula? If there is no $x$, and the formula for quad approx is $Qa(x) = f(a) + f '(a)(x-a) + f ''(a)(x-a)^2/2$, so does it mean that my quadratic approx will be equal just to $f(a)$?
Could anyone help please, it looks easy, but I can't get it. 
 A: When you write $$P(x) = f(a) + f '(a)(x-a) + f ''(a)(x-a)^2/2,$$this polynomial is the approximation near $a$. This means that $P(x)$ will be close to $f(x)$, if $x$ is close to $a$. Another way to express this formula is writing $x - a = h$. So: $$P(a+h) = f(a) + f'(a) h + \frac{f''(a)}{2}h^2.$$This expresses the approximation in terms of the difference $h$, and the closer $h$ is to zero, closer $P(a+h)$ will be to $P(a)$. 
Here your intuition should be clear: we want to approximate $\sqrt[3]{9}$, but the closest integer cube root we know is $\sqrt[3]{8} = 2$. So we want to approximate $\sqrt[3]{x}$ near $8$ ($a = 8$). We want the value of $9$ ($x = 9$). The difference is $1$ ($h = 1$). Now you can use any of the expressions above (pick your favorite).
A: For any $x\in[0,1)$ we have, by considering a Taylor series with Lagrange remainder:
$$ (1+x)^{\frac{1}{3}} = 1+\frac{x}{3}-\frac{x^2}{9}+\frac{5 x^3}{81(1+\xi)^{\frac{8}{3}}},\quad \xi\in(0,x) $$
hence:
$$ \sqrt[3]{1+\frac{1}{8}} = \frac{599}{576}+E,\qquad 0<E<\frac{5}{41472}$$
and:
$$ 0<\sqrt[3]{9}-\frac{599}{288}<\frac{1}{4000}.$$

As a footnote, it is worth mentioning that the Newton's iteration
$$ x \to \frac{3}{x^2}+\frac{2x}{3}$$
with starting point $x=2$ gives $\color{red}{\frac{23401}{11250}}$ as an approximated value for $\sqrt[3]{9}$ in just two steps, and the error in such a case is less than six parts in a million.
A: Perhaps this is what was intended as a quadratic approximation: The Binomial Theorem says
$$
(1+a)^{1/3}=1+\frac a3-\frac{a^2}9+\frac{5a^3}{81}+O(a^4)
$$
which is an alternating series whose terms are decreasing in absolute value for $0\le a\le1$.
Then
$$
\begin{align}
9^{1/3}
&=2\left(1+\frac18\right)^{1/3}\\
&\approx2\left(1+\frac1{24}-\frac1{576}\right)\\[3pt]
&=\frac{599}{288}
\end{align}
$$
which, by the Alternating Series Theorem, is an underestimate by at most the next term, which is $\frac5{20736}$.
A: Since $9=8(1+\frac{1}{8})$, 
$\;\;\sqrt[3]9=2\big(1+\frac{1}{8}\big)^{1/3}=2\big(1+\frac{1}{3}(\frac{1}{8})-\frac{1}{9}(\frac{1}{8})^2+\frac{5}{81}(\frac{1}{8})^3+\cdots\big)$
since $\displaystyle(1+x)^{1/3}=1+\frac{1}{3}x+\frac{(\frac{1}{3})(\frac{1}{3}-1)}{2!}x^2+\frac{(\frac{1}{3})(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!}+\cdots$ 

Since the series is alternating (starting with the second term) and has terms decreasing in absolute value ,
the error in using this approximation is at most $2(\frac{5}{81})(\frac{1}{8})^3$, the absolute value of the first omitted term.
