Taylor polynomial of $(1+x^2)^{1/3}$ Let $f(x)=(1+x^2)^{1/3}$. 
Find the second degree Taylor polynomial $T_2(x)$ of $f(x)$ around $x_0=0$ 
Furthermore, determine a constant $C>0$ such that:
$$ |f(x)-T_2(x)|\leq C|x^3|$$
for all $x \in [-1,1]$

So I have found $T_2(x)$ as such:
$$T_2(x)=\sum_{k=0}^{2} \frac{f^{(k)}x_0}{k!} (x-x_0)^k = 1 + \frac{1}{3}x^2$$
But I don't understand exactly how to determine the constant C. Clearly $ |f(x)-T_2(x)|$ must be the the second degree remainder. I also thought about the range of $f(x)$ and $T_2(x)$ but I don't think that helps in any way. Can someone point me in the right direction?
 A: Hint: Taylor's Theorem with Remainder term has a very specific formula for the remainder term as follows: for a $3$-times continuously differentiable (weaker conditions work but this does the job) real function $f(x)$ on an interval containing $x_0$, we get
$$|R_2(x)| = |f(x) - T_2(x)| = |\frac{f^{(3)}(\xi)}{3!}(x - x_0)^3|$$ for some $\xi \in [x_0, x]$ where $T_2(x)$ is your second degree polynomial. So, in your case,
$$|f(x) - T_2(x)| = \frac{|f^{(3)}(\xi)|}{3!}|x|^3 \leq \max\limits_{s \in [-1, 1]}|f^{(3)}(s)|\frac{|x|^3}{3!}$$ I leave it to you to find the third derivative of $f(x)$ and find an upper bound on $\max\limits_{s \in [-1, 1]}|f^{(3)}(s)|$ (you don't need to compute the maximum exactly; just an upper bound on it will do) to get your full constant. The maximum is guaranteed to exist because $f^{(3)}(s)$ is a continuous function on the compact domain $[1, -1]$.
A: Hint:
You can use that the Taylor series of $\;(1+u)^{1/3} $ is an alternating series. By the binomial formula,
$$(1+u)^{1/3}=1+\tfrac13u-\tfrac 19 u^2+\tfrac 5{81}u^3-\dotsm,$$
hence, by substitution,
$$(1+x^2)^{1/3}=1+\tfrac13 x^2-\tfrac 19 x^4 +\tfrac 5{81}x^6-\dotsm,$$
is also an alternating series. Leibniz' test stipulates that, when approximating  an convergent alternating series by its sum up to rank $n$, the absolute value of the error is no more than the absolute value of the term of rank $n+1$.
