Find 'c' of Lagrange's Mean Value Theorem for the functions :
`f(x)=2x^(2)-1` in the interval [1, 2]

Answer

To find the value of 'c' using Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = 2x^2 - 1 \) on the interval \([1, 2]\), we can follow these steps:

### Step 1: Verify the conditions of LMVT
Lagrange's 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 number \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
For our function \( f(x) = 2x^2 - 1 \):
- It is a polynomial function, hence it is continuous and differentiable everywhere.

### Step 2: Identify the interval
Here, \( a = 1 \) and \( b = 2 \).

### Step 3: Calculate \( f(a) \) and \( f(b) \)
Now we need to calculate \( f(1) \) and \( f(2) \):
\[ f(1) = 2(1)^2 - 1 = 2 \cdot 1 - 1 = 1 \]
\[ f(2) = 2(2)^2 - 1 = 2 \cdot 4 - 1 = 8 - 1 = 7 \]

### Step 4: Calculate the average rate of change
Now we can calculate the average rate of change from \( a \) to \( b \):
\[ \frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(1)}{2 - 1} = \frac{7 - 1}{2 - 1} = \frac{6}{1} = 6 \]

### Step 5: Find the derivative of \( f(x) \)
Next, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(2x^2 - 1) = 4x \]

### Step 6: Set the derivative equal to the average rate of change
According to LMVT, we set \( f'(c) = 6 \):
\[ 4c = 6 \]

### Step 7: Solve for \( c \)
Now, we solve for \( c \):
\[ c = \frac{6}{4} = \frac{3}{2} \]

### Step 8: Verify that \( c \) is in the interval \((1, 2)\)
Since \( c = \frac{3}{2} = 1.5 \) is indeed in the interval \((1, 2)\), we have satisfied the conditions of the theorem.

### Final Answer
Thus, the value of \( c \) is
\[ \boxed{\frac{3}{2}} \]