Solution
= Solution
The cubic Hermite interpolant to the exponential at zero and one is
$$
h(x)=(3-e)x^3+(2e-5)x^2+x+1.
$$
Its endpoint values are one and $e$, and its endpoint derivatives are also one and $e$. The remainder formula gives $e^x-h(x)=e^\xi x^2(x-1)^2/24$ for an interior $\xi$. On this interval $e^\xi\le e$ and $x(1-x)\le1/4$, so
$$
\boxed{|e^x-h(x)|\le\frac e{24}\left(\frac14\right)^2=\frac e{384}}.
$$