1. Let $f(x)=x^{3}+a x+b$, with $a \neq b$, and suppose the tangent lines to the graph of $f$ at $x=a$ and $x=b$ are parallel. Find $f(1)$.
Solution: $\square$
Since $f^{\prime}(x)=3 x^{2}+a$, we must have $3 a^{2}+a=3 b^{2}+a$. Then $a^{2}=b^{2}$, and since $a \neq b$, $a=-b$. Thus $f(1)=1+a+b=1$.