1. (2004 National College Entrance Examination, Tianjin Paper) Given the function $f(x)=a x^{3}+b x^{2}-3 x$ has extreme values at $x= \pm 1$.
( I ) Discuss whether $f(1)$ and $f(-1)$ are the maximum or minimum values of the function $f(x)$;
( II ) Draw a tangent line to the curve $y=f(x)$ through the point $A(0,16)$, find the equation of this tangent line.
1. ( I ) $f^{\prime}(x)=3 a x^{2}+2 b x-3$. From $f^{\prime}(1)=f^{\prime}(-1)=0$, we get $\left\{\begin{array}{l}3 a+2 b-3=0, \\ 3 a-2 b-3=0 .\end{array}\right.$ Solving these, we get $a=1, b=0$.
Thus, $f(x)=x^{3}-3 x, f^{\prime}(x)=3 x^{2}-3=3(x+1)(x-1)$.
Let $f^{\prime}(x)=0$, we get $x=-1$ or $x=1$.
If $x \in(-\infty,-1) \cup(1,+\infty)$, then $f^{\prime}(x)>0$, so $f(x)$ is increasing on $(-\infty, -1)$ and $(1, +\infty)$.
If $x \in(-1,1)$, then $f^{\prime}(x)<0$, so $f(x)$ is decreasing on $(-1,1)$.
Therefore, $f(-1)=2$ is a maximum value, and $f(1)=-2$ is a minimum value.
( II ) The curve equation is $y=x^{3}-3 x$, and point $A(0,16)$ is not on the curve.
Let the tangent point be $M\left(x_{0}, y_{0}\right)$, then $y_{0}=x_{0}^{3}-3 x_{0}$.
Since $f^{\prime}\left(x_{0}\right)=3\left(x_{0}^{2}-1\right)$, the tangent line equation is $y-y_{0}=3\left(x_{0}^{2}-1\right)\left(x-x_{0}\right)$.
Noting that point $A(0,16)$ is on the tangent line, we have $16-\left(x_{0}^{3}-3 x_{0}\right)=3\left(x_{0}^{2}-1\right)\left(0-x_{0}\right)$.
Solving this, we get $x_{0}=-2$.
Thus, the tangent point is $M(-2,-2)$, and the tangent line equation is $9 x-y+16=0$.