The Intermediate Value Theorem (IVT) states that if a function $f(x)$ is continuous on a closed interval $[a, b]$, and if $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c \in (a, b)$ such that $f(c) = 0$.

Step 1: Verify continuity of $f(x)$

The given function is: $f(x) = x^3 + 3x^2 - 9x - 13$ Since $f(x)$ is a polynomial, it is continuous everywhere, including on the interval $[-5, -4]$.

Step 2: Compute $f(a)$ and $f(b)$

Substitute $a = -5$ and $b = -4$ into $f(x)$.

1. Evaluate $f(-5)$: $f(-5) = (-5)^3 + 3(-5)^2 - 9(-5) - 13$ $f(-5) = -125 + 75 + 45 - 13 = -18$

2. Evaluate $f(-4)$: $f(-4) = (-4)^3 + 3(-4)^2 - 9(-4) - 13$ $f(-4) = -64 + 48 + 36 - 13 = 7$

Step 3: Check for opposite signs

We observe: $f(-5) = -18 \quad (\text{negative}), \quad f(-4) = 7 \quad (\text{positive}).$ Since $f(-5)$ and $f(-4)$ have opposite signs, the Intermediate Value Theorem guarantees the existence of a $c \in (-5, -4)$ such that $f(c) = 0$.

Conclusion

By the Intermediate Value Theorem, there is at least one zero of $f(x)$ in the interval $(-5, -4)$.

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Would you like to see a graph of $f(x)$ to further illustrate this? Or do you have additional questions?

Related Questions:

1. How do we find the exact value of $c$ where $f(c) = 0$?
2. Can the Intermediate Value Theorem guarantee more than one zero in $[a, b]$?
3. What are the implications if $f(a)$ and $f(b)$ are both positive (or both negative)?
4. How can we numerically approximate the zero of $f(x)$ in this interval?
5. How does the IVT extend to non-polynomial continuous functions?

Tip:

When applying the Intermediate Value Theorem, always confirm that the function is continuous over the given interval. Discontinuities invalidate the theorem's application.