Suppose $f(x) = x^{6}+3x+1$. In this problem, we will show that $f$ has exactly one root (or zero) in the interval $\lbrack -5, -1 \rbrack$.

(a) First, we show that $f$ has a root in the interval $(-5, -1)$. Since $f$ is a choose continuous differentiable polynomial function on the interval $\lbrack -5, -1 \rbrack$ and $f(-5) =$ and $f(-1) =$ , the graph of $y = f(x)$ must cross the $x$-axis at some point in the interval $(-5, -1)$ by the choose intermediate value theorem mean value theorem squeeze theorem Rolle's theorem . Thus, $f$ has at least one root in the interval $\lbrack -5, -1 \rbrack$.

(b) Second, we show that $f$ cannot have more than one root in the interval $\lbrack -5, -1 \rbrack$ by a thought experiment. Suppose that there were two roots $x = a$ and $x = b$ in the interval $\lbrack -5, -1 \rbrack$ with $a < b$. Then $f(a) = f(b) =$ . Since $f$ is choose continuous differentiable polynomial on the interval $\lbrack -5, -1 \rbrack$ and choose continuous differentiable polynomial on the interval $(-5, -1)$, by choose intermediate value theorem mean value theorem squeeze theorem Rolle's theorem there would exist a point $c$ in interval $(a,b)$ so that $f'(c) = 0$. However, the only solution to $f'(x) = 0$ is $x =$ , which is not in the interval $(a,b)$, since $(a,b) \subseteq \lbrack -5, -1 \rbrack$. Thus, $f$ cannot have more than one root in $\lbrack -5, -1 \rbrack$.

(Note: where the problem asks you to make a choice select the weakest choice that works in the given context. For example "continuous" is a weaker condition than "polynomial" because every polynomial is continuous but not vice-versa. Rolle's theorem is a weaker theorem than the mean value theorem because Rolle's theorem applies to fewer cases.)