https://file.aiquickdraw.com/mathbot/file/1729241847921ah2ese7e.jpg

The question asks whether the **Intermediate Value Theorem (IVT)** applies to the given function \( f(x) = -x^2 + x + 22 \) on the interval \([0, 6]\), and if so, to find \(c\) such that \( f(c) = 2 \). The possible answers involve whether the theorem applies and the value of \(c\).

### To apply the IVT:
1. **Conditions for IVT**: 
   - The function must be continuous on the closed interval \([0, 6]\).
   - If the function is continuous, the IVT states that for any value between \(f(0)\) and \(f(6)\), there exists some \(c \in [0, 6]\) such that \(f(c)\) equals that value.

### Step-by-step verification:
1. **Check continuity**: The function \(f(x) = -x^2 + x + 22\) is a quadratic function, which is continuous everywhere, so it is continuous on \([0, 6]\).

2. **Evaluate \(f(0)\) and \(f(6)\)**:
   - \(f(0) = -(0)^2 + (0) + 22 = 22\)
   - \(f(6) = -(6)^2 + 6 + 22 = -36 + 6 + 22 = -8\)

3. **Check if 2 is between \(f(0)\) and \(f(6)\)**:
   - \(f(0) = 22\) and \(f(6) = -8\). Since \(2\) lies between \(-8\) and \(22\), the IVT guarantees that there is some \(c \in [0, 6]\) such that \(f(c) = 2\).

### Find \(c\) such that \(f(c) = 2\):
Solve \( -x^2 + x + 22 = 2 \):
\[
-x^2 + x + 22 = 2
\]
\[
-x^2 + x + 22 - 2 = 0
\]
\[
-x^2 + x + 20 = 0
\]
Multiply the equation by \(-1\) to simplify:
\[
x^2 - x - 20 = 0
\]
Now solve the quadratic equation using the quadratic formula:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-20)}}{2(1)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 80}}{2}
\]
\[
x = \frac{1 \pm \sqrt{81}}{2}
\]
\[
x = \frac{1 \pm 9}{2}
\]
Thus, \(x = \frac{1 + 9}{2} = 5\) or \(x = \frac{1 - 9}{2} = -4\).

Since \(c\) must be in the interval \([0, 6]\), the solution is \(c = 5\).

### Conclusion:
The Intermediate Value Theorem applies, and the value of \(c\) is \(5\), matching the third option.

---

Do you have any questions or would you like more details?

### Follow-up Questions:
1. How does the Intermediate Value Theorem work for more complex functions?
2. Why is continuity a critical requirement for the IVT?
3. Can the IVT be applied if the function is discontinuous?
4. How would you solve this problem graphically?
5. What other mathematical theorems are closely related to the IVT?

**Tip:** When solving quadratic equations, always check the discriminant to see if real solutions exist before applying the quadratic formula!

Decide whether the Intermediate Value Theorem applies to the given function on the interval [0, 6]: f(x) = -x^2 + x + 22, f(c) = 2.

Learn how to apply the Intermediate Value Theorem to the quadratic function f(x) = -x^2 + x + 22 over the interval [0, 6], with a focus on solving for c when f(c) = 2. Suitable for college-level students studying Calculus I.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation