14. The number $\Delta=\max _{a \leqslant x \leqslant b}|f(x)-g(x)|$ is called the absolute difference of the functions $f(x)$ and $g(x)$ on the closed interval $[a, b]$. Find the absolute difference of the functions $f(x)=x^{2}$ and $g(x)=x^{3}$ on the closed interval $[0,1]$.
14. Since $f(x)-g(x)=x^{2}-x^{3}, f^{\prime}(x)-g^{\prime}(x)=2 x-3 \cdot x^{2}$, by setting $f^{\prime}(x)-g^{\prime}(x)=0$, we get $x=0$ or $\frac{2}{3}$. Also, because $f^{\prime \prime}(x)-g^{\prime \prime}(x)=2-6 x, f^{\prime \prime}\left(\frac{2}{3}\right)-g^{\prime \prime}\left(\frac{2}{3}\right)=2-4=-2<0$, so when $x$ $=\frac{2}{3}$, $f(x)-g(x)$ reaches a maximum value; and since when $0 \leqslant x \leqslant 1$, $f(x)-g(x) \geqslant 0$, the absolute difference $\Delta=f\left(\frac{2}{3}\right)$ $g\left(\frac{2}{3}\right)=\frac{4}{27}$.