Check whether the Lagrange formula is applicable to following functions
`f(x) = 4x^(3) - 5x^(2) +x-2 " on " [0,1]`

Answer

To check whether Lagrange's Mean Value Theorem (LMVT) is applicable to the function \( f(x) = 4x^3 - 5x^2 + x - 2 \) on the interval \([0, 1]\), we need to follow these steps:

### Step 1: Check Continuity
The first condition for applying LMVT is that the function must be continuous on the closed interval \([a, b]\).

**Solution:**
Since \( f(x) \) is a polynomial function, it is continuous everywhere, including the interval \([0, 1]\).

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

**Solution:**
Again, since \( f(x) \) is a polynomial function, it is differentiable everywhere, including the interval \((0, 1)\).

### Step 3: Apply Lagrange's Mean Value Theorem
Since both conditions are satisfied, we can apply LMVT. According to LMVT, there exists at least one point \( c \) in the interval \((0, 1)\) such that:

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

where \( a = 0 \) and \( b = 1 \).

**Solution:**
First, we calculate \( f(0) \) and \( f(1) \):

\[ f(0) = 4(0)^3 - 5(0)^2 + 0 - 2 = -2 \]

\[ f(1) = 4(1)^3 - 5(1)^2 + 1 - 2 = 4 - 5 + 1 - 2 = -2 \]

Now, substituting these values into the LMVT formula:

\[ f'(c) = \frac{f(1) - f(0)}{1 - 0} = \frac{-2 - (-2)}{1 - 0} = \frac{0}{1} = 0 \]

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

**Solution:**
\[ f'(x) = \frac{d}{dx}(4x^3 - 5x^2 + x - 2) = 12x^2 - 10x + 1 \]

### Step 5: Set \( f'(c) = 0 \)
Now, we set \( f'(c) = 0 \):

\[ 12c^2 - 10c + 1 = 0 \]

### Step 6: Solve the Quadratic Equation
To find the value of \( c \), we can use the quadratic formula:

\[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 12 \), \( b = -10 \), and \( c = 1 \).

**Solution:**
\[ c = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 12 \cdot 1}}{2 \cdot 12} \]
\[ = \frac{10 \pm \sqrt{100 - 48}}{24} \]
\[ = \frac{10 \pm \sqrt{52}}{24} \]
\[ = \frac{10 \pm 2\sqrt{13}}{24} = \frac{5 \pm \sqrt{13}}{12} \]

### Conclusion
Since we have found values of \( c \) in the interval \((0, 1)\), we conclude that Lagrange's Mean Value Theorem is applicable to the function \( f(x) = 4x^3 - 5x^2 + x - 2 \) on the interval \([0, 1]\).

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