Let's solve this problem step by step:

Part (a):

We need to show that $f(x) = x^6 + 3x + 1$ has at least one root in the interval $(-3, -1)$.

1. First, observe that $f(x)$ is a polynomial function, and every polynomial is continuous. Thus, $f(x)$ is continuous on the interval $[-3, -1]$.

2. Now, we evaluate the function at the endpoints of the interval: $f(-3) = (-3)^6 + 3(-3) + 1 = 729 - 9 + 1 = 721$ $f(-1) = (-1)^6 + 3(-1) + 1 = 1 - 3 + 1 = -1$

3. Since $f(-3) = 721 > 0$ and $f(-1) = -1 < 0$, the function takes opposite signs at $x = -3$ and $x = -1$.

4. By the intermediate value theorem (IVT), which states that if a function is continuous on an interval and takes opposite signs at the endpoints, there must be at least one root between the endpoints, there exists a point $c \in (-3, -1)$ such that $f(c) = 0$.

Thus, $f(x)$ has at least one root in the interval $(-3, -1)$.

Part (b):

We now show that $f(x)$ cannot have more than one root in the interval $[-3, -1]$.

1. Suppose, for the sake of contradiction, that there are two distinct roots $a$ and $b$ in $[-3, -1]$, where $a < b$ and $f(a) = f(b) = 0$.

2. Since $f(x)$ is a polynomial function, it is also differentiable on $[-3, -1]$.

3. By Rolle's theorem (a special case of the mean value theorem), which applies when a function is continuous and differentiable on an interval and takes the same value at two distinct points, there must be a point $c \in (a, b)$ such that $f'(c) = 0$.

4. Now, compute the derivative of $f(x)$: $f'(x) = 6x^5 + 3$ We want to find where $f'(x) = 0$: $6x^5 + 3 = 0 \implies 6x^5 = -3 \implies x^5 = -\frac{1}{2}$ Taking the fifth root of both sides: $x = \left( -\frac{1}{2} \right)^{1/5}$ This solution is a single number, but it does not lie in the interval $(-3, -1)$. Thus, there is no point in $(a, b) \subseteq (-3, -1)$ where $f'(x) = 0$.

Therefore, by Rolle's theorem, $f(x)$ cannot have more than one root in the interval $[-3, -1]$.

Final Conclusion:

• In part (a), we showed that $f(x)$ has at least one root in $(-3, -1)$ by the intermediate value theorem.
• In part (b), we showed that $f(x)$ cannot have more than one root in $[-3, -1]$ by Rolle's theorem.

Thus, $f(x)$ has exactly one root in the interval $[-3, -1]$.

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Would you like more details on any step or have any questions?

Here are five related questions:

1. What are the conditions for applying the intermediate value theorem?
2. How do you prove Rolle's theorem geometrically?
3. Can you generalize this result to other intervals or polynomials?
4. How does the behavior of higher-order derivatives affect the shape of polynomial functions?
5. What are the implications if $f'(x)$ had more than one critical point?

Tip: Understanding the mean value theorem and Rolle's theorem is crucial for proving the uniqueness of roots in certain intervals.