2. [3] Determine the real number $a$ having the property that $f(a)=a$ is a relative minimum of $f(x)=$ $x^{4}-x^{3}-x^{2}+a x+1$.
Answer: 1 . Being a relative minimum, we have $0=f^{\prime}(a)=4 a^{3}-3 a^{2}-2 a+a=a(4 a+1)(a-1)$. Then $a=0,1,-1 / 4$ are the only possibilities. However, it is easily seen that $a=1$ is the only value satisfying $f(a)=a$.