Solution
= Solution
\b[Not <injective>; <surjective>.] Indeed, $f(0)=f(-1)=-9$ although $0\ne-1$. This <polynomial> is <continuous>, with $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to\infty$ as $x\to\infty$. For any prescribed real $y$, choose $a<b$ with $f(a)<y<f(b)$; the <intermediate value theorem> supplies a <preimage> between $a$ and $b$.