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?

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 $$[1, -1]$$.
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 $$n+1$$.