= Solution
For $f\notin E$ and rational $q\in(0,1)$, every $M$ admits $m\geq M$ with
$$
\left|\frac{f^{(m)}(q)}{m!}\right|>m^m.
$$
Consequently
$$
\limsup_{m\to\infty}
\left|\frac{f^{(m)}(q)}{m!}\right|^{1/m}=\infty.
$$
By the <Cauchy-Hadamard theorem>, the <Taylor series> of $f$ at $q$ has <radius of convergence> zero.
Suppose some Taylor series represented $f$ throughout a neighborhood of a point $x\in[0,1]$. That neighborhood contains a rational $q\in(0,1)$. A function represented locally by a convergent power series is a <real analytic function>, and re-expanding that series about $q$ gives the Taylor series of $f$ at $q$ a positive radius of convergence. This contradicts the preceding conclusion. Thus $f$ agrees with no Taylor series on a neighborhood of any point; it is a <nowhere-analytic generic smooth function>.
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