Verify the conditions of Mean Value Theorem in the following. In each case, find a point in the interval as stated by the Mean Value Theorem :
`f(x)=ax^(2)+bx+c" on "[0,1]`

Answer

To verify the conditions of the Mean Value Theorem (MVT) for the function \( f(x) = ax^2 + bx + c \) on the interval \([0, 1]\), we will follow these steps:

### Step 1: Check Continuity
The first condition of the Mean Value Theorem states that the function must be continuous on the closed interval \([a, b]\).

**Solution:**
The function \( f(x) = ax^2 + bx + c \) is a polynomial function. Polynomial functions are continuous everywhere on the real number line, including the closed interval \([0, 1]\). Therefore, \( f(x) \) is continuous on \([0, 1]\).

### Step 2: Check Differentiability
The second condition of the Mean Value Theorem states that the function must be differentiable on the open interval \((a, b)\).

**Solution:**
Since \( f(x) \) is a polynomial, it is also differentiable everywhere on the real number line. Thus, \( f(x) \) is differentiable on the open interval \((0, 1)\).

### Step 3: Apply the Mean Value Theorem
Now that we have verified the continuity and differentiability, we can apply the Mean Value Theorem. According to MVT, there exists at least one point \( c \) in the interval \((0, 1)\) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

For our case, \( a = 0 \) and \( b = 1 \).

**Solution:**
1. Calculate \( f(0) \) and \( f(1) \):
- \( f(0) = a(0)^2 + b(0) + c = c \)
- \( f(1) = a(1)^2 + b(1) + c = a + b + c \)

2. Now, compute the right-hand side of the MVT:
\[ \frac{f(1) - f(0)}{1 - 0} = \frac{(a + b + c) - c}{1} = a + b \]

3. Now, find the derivative \( f'(x) \):
\[ f'(x) = 2ax + b \]

4. Set \( f'(c) = a + b \):
\[ 2ac + b = a + b \]

5. Simplifying gives:
\[ 2ac = a \implies c = \frac{a}{2a} = \frac{1}{2} \quad (\text{assuming } a \neq 0) \]

### Conclusion
Thus, we have found that there exists a point \( c = \frac{1}{2} \) in the interval \((0, 1)\) that satisfies the conditions of the Mean Value Theorem.

### Summary of Conditions Verified
1. **Continuity**: \( f(x) \) is continuous on \([0, 1]\).
2. **Differentiability**: \( f(x) \) is differentiable on \((0, 1)\).
3. **Existence of \( c \)**: There exists \( c = \frac{1}{2} \) such that \( f'(c) = a + b \).