= Solution
The global bound requires
$$
\frac1{180n^4}<10^{-10}
\quad\Longleftrightarrow\quad
n>\left(\frac{10^{10}}{180}\right)^{1/4}\approx86.3340.
$$
Since the strip count must be even, \b[the smallest count certified by this bound is $n=88$].
For the tighter bound, use the <bisection method> on the continuous extension $B(x)-10^{-10}$ for $x\ge2$. The initial <root bracket> $[2,88]$ is valid: $B(2)=1/2880>10^{-10}$, while $B(88)$ is below the global bound and hence below $10^{-10}$. To identify the smallest admissible even count, the monotonicity of $B$ can also be justified.
Put $h=2/x$ and
$$
C(h)=(e-1)\frac h{e^h-1}-\frac e2+\frac{eh}{2},
\qquad B(x)=\frac{h^4C(h)}{2880}.
$$
For $h>0$, differentiation shows
$$
\left(\frac h{e^h-1}\right)'+\frac12
=\frac{e^{2h}-1-2he^h}{2(e^h-1)^2}\ge0.
$$
The numerator is nonnegative because $e^h-e^{-h}\ge2h$: its difference from $2h$ starts at zero and has nonnegative <derivative> $e^h+e^{-h}-2$, by the <arithmetic-geometric mean inequality>. Thus $C'(h)\ge1/2$, and $C(0^+)=e/2-1>0$. Therefore $h^4C(h)$ increases with $h$, so $B(x)$ is strictly decreasing with $x$. <Bisection> locates its unique threshold crossing; then check the neighboring even integers, retaining the strict error inequality.
For example, $B(66)\approx1.09626\times10^{-10}$ and $B(68)\approx9.71694\times10^{-11}$, so \b[the tighter bound certifies $68$ strips]. These are evaluations of the bound, not an assumption that it equals the actual <quadrature error>.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-simpson-error.png]
{title=Computed Simpson errors and the two certified bounds, with the required error level}
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