Taylor polynomial for $f(x)=x^{1/7}$ about $a=1$ Here is the problem:

(a) Determine the Taylor polynomial $T_2(x)$ of degree $2$ for the function $f(x)=x^{1/7}$ centered at $a=1$.
(b) Suppose we were to use the approximation $f(x) \approx T_2(x)$. Obtain an upper bound on the error of this approximation when $0.7 \leq x \leq 1.3$. Give your answer rounded (up) to $4$ decimal places.

I'm stuck on (a) currently, I can't figure out the formula to use to get a Taylor series of a certain degree.
 A: To get your polynomial you need to fill in the values of the second degree Taylor polynomial:
$$P\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) + f''\left( a \right)\frac{{{{\left( {x - a} \right)}^2}}}{{2!}}$$
In general, the Taylor polynomial of $n$th degree around $x=a$ is given by
$${P_n}\left( x \right) = f\left( a \right) + f'\left( a \right)\frac{{x - a}}{{1!}} +  \cdots  + {f^{\left( {n - 1} \right)}}\left( a \right)\frac{{{{\left( {x - a} \right)}^{n - 1}}}}{{\left( {n - 1} \right)!}} + {f^{\left( n \right)}}\left( a \right)\frac{{{{\left( {x - a} \right)}^n}}}{{n!}}$$
At $x=1$ we have
$f(1) = 1$
$f'(1) = \dfrac{1}{7}$
$f''(1) = -\dfrac{6}{49}$
So we replace
$$P\left( x \right) = 1 + \dfrac{1}{7}\left( {x - 1} \right) -\dfrac{6}{49}\frac{{{{\left( {x - 1} \right)}^2}}}{{2!}}$$
You can see the respective plots here:

For the error, we calculate the remainder of the polynomial, which is given by several formulas, and among one is:
$$R{_n}\left( x \right) = \int\limits_a^x {\frac{{{{\left( {x - t} \right)}^n}}}{{n!}}{f^{\left( {n + 1} \right)}}\left( t \right)dt} $$
If you're not familiar with integration you can use the following theorem:
Suppose $m \leq f^{(n+1)}(x) \leq M$ in some interval $I$ containing $x=a$. Then the error or remainder of degree $n$, $R_n(x)$ satisfies:
$$m\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}} \leq {R_n}\left( x \right) \leq M\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}}$$
If $x < a$ the error as to be multiplied by $(-1)^{n+1}$, i.e
$$m\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}} \leq (-1)^{n+1}{R_n}\left( x \right)  \leq M\frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}}$$
In your case you have $0.7 < x < 1.3$ so you can use the theorem and get using $M$ and $m$ as $f'''(0.7)$ and $f'''(1.3)$ since the derivative is monotonically decreasing.
$$0.11\frac{{{{\left( {x - 1} \right)}^3}}}{{3!}} < {R_2}\left( x \right) < 0.63\frac{{{{\left( {x - 1} \right)}^3}}}{{3!}}$$
So using this you can calculate what the error is for example for $x=1.3$
$$0.000495 < {R_2}\left( {1.3} \right) < 0.002835$$
You can see the plots of the cubics here (red and blue), which give upper and lower bounds of the error (green) for $x$ near $1$ (and obviously give $0$ when $x=1$).

