If $p(x) = x^4-4x^3+6x^2-4x+1$ is the Taylor polynomial of $f$ around $x=1$, then $1$ is a local minimum 
Consider $f:\mathbb{R} \to \mathbb{R} \in C^4$. Show that $p(x) =
x^4-4x^3+6x^2-4x+1$ is the Taylor polynomial of order $4$ of $f$ around $x=1$, then
  $1$ is a local minimum.

I'm not sure how to proceed. I know that $p(x) = \sum\limits_{k=0}^{4} \dfrac{f^{(n)}(1)(x-1)^n}{n!} = f(1)+f'(1)(x-1)+\dfrac{f''(1)(x-1)^2}{2} + \dfrac{f'''(1)(x-1)^{3}}{6} + \dfrac{f^{(4)}(1)(x-1)^4}{24}=x^4-4x^3+6x^2-4x+1$ 
and I can see that $p(1) = 0 \implies f(1)=0$. But how can I obtain information about $f'(1)$ and $f''(1)$?

By the Remainder theorem of the Taylor polynomials we have $$ f(x)=P(x) + \frac {f^{(5)}(\xi)}{5!} (x-1)^5$$
$$ = (x-1)^4 + \frac {f^{(5)}(\xi)}{5!} (x-1)^5$$ 
$$ = (x-1)^4 \big[1+\frac {f^{(5)}(\xi)}{5!} (x-1)\big]$$
Let $x$ be close enough to $1$ such that $$ \big| \frac {f^{(5)}(\xi)}{5!} (x-1)\big|<1/2$$
Then $$ \frac {(x-1)^4}{2}<f(x)<\frac{3(x-1)^4}{2}$$
Thus $f(x)$ has a minimum at $x=1$.
