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)+ex+e" on "[0,1]`

Answer

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

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

1. **Continuity**: Since \( f(x) = ax^2 + ex + e \) is a polynomial function, it is continuous everywhere, including on the interval \([0, 1]\).
2. **Differentiability**: Polynomial functions are also differentiable everywhere. Therefore, \( f(x) \) is differentiable on the interval \((0, 1)\).

### Step 2: Calculate \( f(0) \) and \( f(1) \)
Next, we need to evaluate the function at the endpoints of the interval:

- \( f(0) = a(0)^2 + e(0) + e = e \)
- \( f(1) = a(1)^2 + e(1) + e = a + e + e = a + 2e \)

### Step 3: Apply the Mean Value Theorem
According to the Mean Value Theorem, there exists at least one point \( c \) in the interval \((0, 1)\) such that:

\[ f'(c) = \frac{f(1) - f(0)}{1 - 0} \]

Calculating the right-hand side:

\[ f'(c) = \frac{(a + 2e) - e}{1} = a + e \]

### Step 4: Find the Derivative \( f'(x) \)
Now, we need to find the derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx}(ax^2 + ex + e) = 2ax + e \]

### Step 5: Set Up the Equation
We set \( f'(c) = a + e \):

\[ 2ac + e = a + e \]

### Step 6: Solve for \( c \)
Now, we can simplify the equation:

\[ 2ac = a \]

Assuming \( a \neq 0 \):

\[ c = \frac{1}{2} \]

### Conclusion
Thus, we have verified the conditions of the Mean Value Theorem, and we found that the point \( c \) in the interval \((0, 1)\) is:

\[ c = \frac{1}{2} \]