Let $f: [0,1] \to \mathbb{R}$ be a function and suppose that $f$ is differentiable on $[0,1]$. Suppose that $f'(0)=f''(0)=\ldots=f^{(5)}(0)=0$ and $f^{(6)}(0)>0$. Show that there exists a $\epsilon >0$ such that $ f(x) \ge f(0)+\displaystyle\frac{f^{(6)}(0)}{721}x^6$ for all $x \in [0,\epsilon)$
Here is what I have got so far, using remainder theorem of Taylor polynomial we have $f(x) - \left(f(0)+\displaystyle\frac{f^{(6)}(0)}{721}x^6\right)=\displaystyle\frac{f^{(7)}(0)}{7!}x^7$ for $ 0 \le c \le x$. The problem is that I need to show $f^{(7)}(c) \ge 0$. This is where I got stuck. Any hint would be greatly appreciated.