We are tasked with using the Intermediate Value Theorem (IVT) to determine if the function $f(x) = x^2 - 3$ has a zero in the interval $[-4, 4]$.

Step-by-Step Analysis:

1. Check Continuity: The function $f(x) = x^2 - 3$ is a polynomial, and all polynomials are continuous over their entire domains. Thus, $f(x)$ is continuous on the interval $[-4, 4]$.

2. Evaluate the Function at Key Points: To apply the IVT, we need to find $f(x)$ at two points in the interval $[-4, 4]$ such that the function values have opposite signs (i.e., one positive and one negative).

   Let's compute $f(x)$ at a few key points:

   • $f(-4) = (-4)^2 - 3 = 16 - 3 = 13$
   • $f(-1) = (-1)^2 - 3 = 1 - 3 = -2$
   • $f(0) = (0)^2 - 3 = -3$
   • $f(4) = (4)^2 - 3 = 16 - 3 = 13$

   Now we can see that:

   • $f(-1) = -2$ (negative)
   • $f(4) = 13$ (positive)

3. Application of IVT: By the Intermediate Value Theorem, if $f(x)$ is continuous on $[-1, 4]$, and $f(-1) = -2$ and $f(4) = 13$, there must exist some $c \in (-1, 4)$ such that $f(c) = 0$. This confirms the existence of a zero in this interval.

Correct Option:

• The first option is correct: $f(x)$ is continuous on $[-4, 4]$, $f(-1) = -2$, $f(4) = 13$, so we can use the IVT to show there is a zero between $[-1, 4]$.

Would you like more details or have any questions about this?

Relative Questions:

1. How does the IVT ensure a zero exists in the interval?
2. What happens if $f(x)$ values at both points are positive?
3. How can we verify the exact zero of the function?
4. What other theorems can be used for finding roots of functions?
5. How does continuity play a role in applying the IVT?

Tip:

The Intermediate Value Theorem is only applicable if the function is continuous over the given interval. Always check this condition first!