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The question asks you to use the Intermediate Value Theorem (IVT) to show that the function \( P(x) = 4x^4 - 4x^2 + 10x - 8 \) has a zero between 0.5 and 1.

### Steps:
1. **Evaluate the function at the lower bound (x = 0.5):**
   \[
   P(0.5) = 4(0.5)^4 - 4(0.5)^2 + 10(0.5) - 8
   = 4(0.0625) - 4(0.25) + 5 - 8
   = 0.25 - 1 + 5 - 8
   = -3.75
   \]
   So, \( P(0.5) = -3.75 \).

2. **Evaluate the function at the upper bound (x = 1):**
   \[
   P(1) = 4(1)^4 - 4(1)^2 + 10(1) - 8
   = 4 - 4 + 10 - 8
   = 2
   \]
   So, \( P(1) = 2 \).

### Applying the Intermediate Value Theorem (IVT):
- The IVT states that if a function is continuous on the interval \([a, b]\), and the function takes values of opposite signs at the endpoints \(a\) and \(b\), then there exists at least one \(c\) in the interval \((a, b)\) such that \(P(c) = 0\).
- Here, \( P(0.5) = -3.75 \) and \( P(1) = 2 \), so the function changes sign between \(x = 0.5\) and \(x = 1\).
- Since \(P(x)\) is a polynomial (and hence continuous), by the IVT, there is at least one zero between 0.5 and 1.

Let me know if you'd like more details or have any other questions!  

### Further Exploration:
1. What does the IVT tell us about the behavior of a function over an interval?
2. How can you confirm if a polynomial function is continuous?
3. Could the function \( P(x) \) have more than one zero between 0.5 and 1? Why or why not?
4. How would the derivative of \( P(x) \) help in locating the exact zero?
5. What are some real-world applications of the Intermediate Value Theorem?

**Tip**: When working with polynomials, remember that they are continuous and differentiable everywhere!

Use the intermediate value theorem to show that the function has a zero between the two numbers 0.5 and 1 for P(x) = 4x^4 - 4x^2 + 10x - 8.

Learn how to apply the Intermediate Value Theorem to find a zero of the polynomial P(x) = 4x^4 - 4x^2 + 10x - 8 between 0.5 and 1. Step-by-step calculations included, along with explanations of the IVT.

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