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.