To solve the problem, we need to analyze the given information about the function \( f(x) \) and apply Rolle's Theorem and the Mean Value Theorem.
### Step-by-Step Solution:
1. **Given Information**:
We know that:
\[
f(1) = 2, \quad f(2) = 5, \quad f(3) = 10
\]
2. **Define a New Function**:
Let's define a new function:
\[
g(x) = f(x) - (x^2 + 1)
\]
This function will help us analyze the behavior of \( f(x) \).
3. **Evaluate \( g(x) \) at the Given Points**:
Now we will evaluate \( g(x) \) at the points 1, 2, and 3:
- For \( x = 1 \):
\[
g(1) = f(1) - (1^2 + 1) = 2 - 2 = 0
\]
- For \( x = 2 \):
\[
g(2) = f(2) - (2^2 + 1) = 5 - 5 = 0
\]
- For \( x = 3 \):
\[
g(3) = f(3) - (3^2 + 1) = 10 - 10 = 0
\]
4. **Apply Rolle's Theorem**:
Since \( g(1) = g(2) = g(3) = 0 \), we can apply Rolle's Theorem. According to Rolle's Theorem, since \( g(x) \) is continuous and differentiable, there exists at least one \( c_1 \) in the interval \( (1, 2) \) such that:
\[
g'(c_1) = 0
\]
5. **Find the Derivative of \( g(x) \)**:
Now we find the derivative of \( g(x) \):
\[
g'(x) = f'(x) - 2x
\]
Setting \( g'(c_1) = 0 \):
\[
f'(c_1) - 2c_1 = 0 \implies f'(c_1) = 2c_1
\]
6. **Apply Rolle's Theorem Again**:
Now, since \( g(2) = 0 \) and \( g(3) = 0 \), we can apply Rolle's Theorem again in the interval \( (2, 3) \). There exists at least one \( c_2 \) in \( (2, 3) \) such that:
\[
g'(c_2) = 0
\]
This gives us:
\[
f'(c_2) - 2c_2 = 0 \implies f'(c_2) = 2c_2
\]
7. **Use the Mean Value Theorem**:
Now, we can apply the Mean Value Theorem to \( f'(x) \) in the interval \( (c_1, c_2) \):
\[
f''(c_3) = \frac{f'(c_2) - f'(c_1)}{c_2 - c_1}
\]
Since \( f'(c_1) = 2c_1 \) and \( f'(c_2) = 2c_2 \), we have:
\[
f''(c_3) = \frac{2c_2 - 2c_1}{c_2 - c_1}
\]
This simplifies to:
\[
f''(c_3) = 2
\]
8. **Conclusion**:
Therefore, we conclude that there exists some \( c_3 \) in \( (1, 3) \) such that:
\[
f''(c_3) = 2
\]
### Final Answer:
Thus, the result we find is that \( f''(c) = 2 \) for some \( c \) in the interval \( (1, 3) \).
