Verify mean value theorem for the function `f(x) = x^(3)-2x^(2)-x+3` in `[0,1]`

Explanation

To verify the Mean Value Theorem (MVT) for the function \( f(x) = x^3 - 2x^2 - x + 3 \) in the interval \([0, 1]\), we will follow these steps:

### Step 1: Check the conditions of the Mean Value Theorem
The Mean Value Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that:

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

In our case, \( f(x) \) is a polynomial function, which is continuous and differentiable everywhere. Therefore, it satisfies the conditions of the MVT on the interval \([0, 1]\).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Let \( a = 0 \) and \( b = 1 \).

\[ f(0) = 0^3 - 2(0^2) - 0 + 3 = 3 \]
\[ f(1) = 1^3 - 2(1^2) - 1 + 3 = 1 - 2 - 1 + 3 = 1 \]

### Step 3: Compute \( f(b) - f(a) \)
Now, we calculate:

\[ f(b) - f(a) = f(1) - f(0) = 1 - 3 = -2 \]

### Step 4: Compute \( b - a \)
Next, we find:

\[ b - a = 1 - 0 = 1 \]

### Step 5: Calculate the average rate of change
Now we can find the average rate of change of the function on the interval \([0, 1]\):

\[ \frac{f(b) - f(a)}{b - a} = \frac{-2}{1} = -2 \]

### Step 6: Find the derivative \( f'(x) \)
Now we need to find the derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx}(x^3 - 2x^2 - x + 3) = 3x^2 - 4x - 1 \]

### Step 7: Set \( f'(c) \) equal to the average rate of change
According to the MVT, we need to solve for \( c \) in the equation:

\[ f'(c) = -2 \]

This gives us:

\[ 3c^2 - 4c - 1 = -2 \]

### Step 8: Rearranging the equation
Rearranging the equation, we get:

\[ 3c^2 - 4c + 1 = 0 \]

### Step 9: Solve the quadratic equation
Now we will solve the quadratic equation \( 3c^2 - 4c + 1 = 0 \) using the quadratic formula:

\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \]

Calculating the discriminant:

\[ = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm \sqrt{4}}{6} = \frac{4 \pm 2}{6} \]

Calculating the two possible values for \( c \):

\[ c_1 = \frac{6}{6} = 1, \quad c_2 = \frac{2}{6} = \frac{1}{3} \]

### Step 10: Verify the values of \( c \)
Both \( c = 1 \) and \( c = \frac{1}{3} \) lie within the interval \((0, 1)\). Thus, the Mean Value Theorem is verified for the function \( f(x) \) on the interval \([0, 1]\).

### Summary
We have verified that the Mean Value Theorem holds for the function \( f(x) = x^3 - 2x^2 - x + 3 \) in the interval \([0, 1]\).

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