# Question about a solution to a problem involving Taylor's theorem and local minimum

I've been studying "Berkeley Problems in Mathematics, Souza, Silva" and I came across this problem:

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a $C^{\infty}$ function. Assume that $f(x)$ has a local minimum at $x = 0$. Prove there is a disc centered on the $y$ axis which lies above the graph of $f$ and touches the graph at $(0, f(0))$.

We use Taylor's theorem:

there is a constant $C$ such that $|f(x) - f(0) - f’(0)x| \le Cx^2$ and we assume that $|x| < 1$.

Why is that?

I know that if a function has a local minimum at $0$, it means that in a certain neighbourhood its values cannot be less than $f(0)$.

Will anything bad happen if we instead assume that $|x|<\delta<1$ ?

Please help me. I see it's a crucial step in the solution of this problem.

question: Problem 1.4.26 page 24 , solution: page 177

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Suppose that $|f''(x)|\le M$ for $|x|\le\epsilon$. Since $f'(0)=0$, we have that for $|x|\lt\epsilon$, $$\left|\frac{f'(x)-f'(0)}{x-0}\right|=|f''(\xi)|\le M$$ for some $\xi$ between $0$ and $x$; that is, for $|\xi|\le\epsilon$.

Thus, $|f'(x)|\le M|x|$ and integrating yields $|f(x)-f(0)|\le\frac M2x^2$ for $|x|\le\epsilon$.

The bottom of the circle of radius $r$ centered at $f(0)+r$ is $$g(x)=f(0)+r-\sqrt{r^2-x^2}$$ so that $g(0)=f(0)$ and \begin{align} g(x)-f(0) &=\frac{x^2}{r+\sqrt{r^2-x^2}}\\ &\ge\frac1{2r}x^2 \end{align} Thus, if we set $r\le\min\left(\frac1M,\epsilon\right)$, the circle of radius $r$ centered at $f(0)+r$ satisfies the requirements. That is, $$g(x)\ge f(0)+\frac1{2r}x^2\ge f(0)+\frac M2x^2\ge f(x)\\ g(0)=f(0)$$ and the circle only extends to $|x|\le r\le\epsilon$.

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