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The problem asks to determine whether the **Intermediate Value Theorem (IVT)** applies to the given function on the interval \([0,6]\) and whether \( f(c) = 2 \) has a solution in that interval.

### Step-by-step breakdown:

1. **Given Function:**
   \[
   f(x) = -x^2 + x + 22
   \]
   The function is a polynomial, which is continuous everywhere, including the interval \([0,6]\). Therefore, the continuity condition of the IVT is satisfied.

2. **Interval and values:**
   We need to check if \( f(0) \) and \( f(6) \) bracket the value 2. 

   - Calculate \( f(0) \):
     \[
     f(0) = -(0)^2 + 0 + 22 = 22
     \]
   - Calculate \( f(6) \):
     \[
     f(6) = -(6)^2 + 6 + 22 = -36 + 6 + 22 = -8
     \]

   So, \( f(0) = 22 \) and \( f(6) = -8 \). 

3. **IVT condition:**
   The IVT guarantees that for any value \( L \) between \( f(0) = 22 \) and \( f(6) = -8 \), there exists a \( c \in (0,6) \) such that \( f(c) = L \).

   Since \( 2 \) is between \( 22 \) and \( -8 \), there must be some \( c \in (0,6) \) such that \( f(c) = 2 \).

4. **Finding \( c \) where \( f(c) = 2 \):**
   We need to solve the equation \( f(x) = 2 \):

   \[
   -x^2 + x + 22 = 2
   \]
   Simplifying this:
   \[
   -x^2 + x + 22 - 2 = 0
   \]
   \[
   -x^2 + x + 20 = 0
   \]
   Multiplying through by \(-1\):
   \[
   x^2 - x - 20 = 0
   \]
   Solving this quadratic equation using the quadratic formula:
   \[
   x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-20)}}{2(1)}
   \]
   \[
   x = \frac{1 \pm \sqrt{1 + 80}}{2}
   \]
   \[
   x = \frac{1 \pm \sqrt{81}}{2}
   \]
   \[
   x = \frac{1 \pm 9}{2}
   \]
   So:
   \[
   x = \frac{1 + 9}{2} = 5 \quad \text{or} \quad x = \frac{1 - 9}{2} = -4
   \]
   Since \( x = -4 \) is outside the interval \([0,6]\), the valid solution is \( x = 5 \).

### Conclusion:
The IVT applies, and the value of \( c \) where \( f(c) = 2 \) is \( c = 5 \).

The correct answer is:
\[
\boxed{\text{Yes, the Intermediate Value Theorem applies and } c = 5.}
\]

Would you like any more details on this explanation or another concept? Here are some related questions to explore:

1. Why is the Intermediate Value Theorem important in calculus?
2. How can we generalize the application of the IVT to other types of functions?
3. What happens if a function is not continuous on the interval?
4. Can the IVT be applied if \( f(0) \) and \( f(6) \) don't bracket the target value?
5. How does solving a quadratic equation differ if its discriminant is negative?

**Tip:** When applying the IVT, always ensure the function is continuous over the interval you're considering.

Decide whether the Intermediate Value Theorem applies to the given function on the indicated interval. If so, find c as guaranteed by the theorem.

Learn how to apply the Intermediate Value Theorem to a continuous polynomial function, solving for the value of c where f(c) = 2 on the interval [0, 6]. This calculus problem involves solving a quadratic equation using the quadratic formula.

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