6.72. $f(x)=x^{3}-3 x^{2}+1 ; x \in(-1 ; 1)$.
Solution. $f^{\prime}(x)=3 x^{2}-6 x=3 x \cdot(x-2)=0$ at $x_{1}=0$, $x_{2}=2$. The interval $x \in(-1 ; 1)$ only includes $x_{1}=0$. We have for $x \in(-1 ; 0) f^{\prime}(x)>0$, and the function is increasing, for $x \in(0 ; 1) f^{\prime}(x)<0$, and the function is decreasing, on the interval $x \in(-1 ; 1)$ there is only a global maximum $f_{\max }\left(x_{\max }=\right.$ $=0$ ) $=1$, the minimum value does not exist.