6.69. $f(x)=2 x^{3}-3 x^{2}-12 x-1$.

6.69. $f(x)=2 x^{3}-3 x^{2}-12 x-1$.
The solution is as follows. We find $f^{\prime}(x)=6 x^{2}-6 x-12=6(x+1)(x-2)$. $f^{\prime}(x)=0$ when $x=-1$ and when $x=2$. We have for $x \in(-\infty ;-1) \cup$ $\cup(2 ;+\infty) f^{\prime}(x)>0$, and on these intervals the function is increasing, for $x \in(-1 ; 2) f^{\prime}(x)<0$, and here the function is decreasing. Since the first derivative changes its sign from “+” to “-” when passing through the first kind critical point $x=-1$, there is a local maximum at this point $f_{\max }(-1)=6$. When passing through the other first kind critical point $x=2$, the first derivative changes its sign from “-” to "+", therefore, there is a local minimum at this point $f_{\text {min }}(2)=-29$.

PROOF OF THE UNIQUENESS OF THE SOLUTION OF THE EQUATION $f(x)=a$

FOR ANY VALUE OF THE PARAMETER $a$ USING THE PRINCIPLE OF MONOTONICITY