I want to make sure I'm doing this right.

Find the third degree Taylor polynomial of $\displaystyle f(x) = \cos x$ centered at c $\displaystyle = $ $\displaystyle \frac {\pi}{6}$. Write the remainder in Lagrange form and.find an upper bound for the remainder

So $\displaystyle n = 3$ and $\displaystyle c = \frac{\pi}{6}$ right?

$\displaystyle f(x) = \cos x$; $\displaystyle f{(\frac{\pi}{6}}) = \frac{\sqrt{3}}{2}$

$\displaystyle f'(x) = -\sin x$; $\displaystyle f'{(\frac{\pi}{6}}) = -\frac{1}{2}$

$\displaystyle f''(x) = -\cos x$; $\displaystyle f''{(\frac{\pi}{6}}) = -\frac{\sqrt{3}}{2}$

$\displaystyle f'''(x) = \sin x$; $\displaystyle f'''{(\frac{\pi}{6}}) = \frac{1}{2}$

(error) $\displaystyle f^{(4)}(x) = \cos x$;

$\displaystyle P_3(\frac{\pi}{6}) = (\frac{\sqrt{3}}{2}) - (\frac{1}{2 * 1!} (x - \frac{\pi}{6})) - (\frac{\sqrt{3}}{2 * 2!}(x - \frac{\pi}{6})^2) + (\frac{1}{2 * 3!}(x - \frac{\pi}{6})^3) $

$\displaystyle R_3(x) = \frac{cos (c)}{4!}(x - \frac{\pi}{6})^4$, where $\displaystyle c \exists {[\frac{\pi}{6}, x]}$

I was confused about the part where it says to find an upper bound for the remainder. Does $\displaystyle c \exists {(\frac{\pi}{6}, x)}$ show the upper bound or do I need more?