Find 'c' of Lagrange's Mean Value Theorem for the functions :
`f(x)=e^(x)` in the interval [0, 1]

Answer

To find the value of \( c \) using Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = e^x \) in the interval \([0, 1]\), we will 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 \( c \) in \((a, b)\) such that:

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

In our case, \( f(x) = e^x \), which is continuous and differentiable everywhere. Therefore, it satisfies the conditions of LMVT on the interval \([0, 1]\).

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

\[ f(0) = e^0 = 1 \]
\[ f(1) = e^1 = e \]

### Step 3: Apply LMVT
Now, we can calculate the right-hand side of the equation:

\[ \frac{f(b) - f(a)}{b - a} = \frac{f(1) - f(0)}{1 - 0} = \frac{e - 1}{1} = e - 1 \]

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

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

### Step 5: Set up the equation
According to LMVT, we need to find \( c \) such that:

\[ f'(c) = e^c = e - 1 \]

### Step 6: Solve for \( c \)
To find \( c \), we need to solve the equation:

\[ e^c = e - 1 \]

Taking the natural logarithm on both sides:

\[ c = \ln(e - 1) \]

### Step 7: Determine the interval for \( c \)
Since \( e \) is approximately \( 2.718 \), we can compute \( e - 1 \):

\[ e - 1 \approx 1.718 \]

Thus,

\[ c = \ln(1.718) \]

Calculating \( \ln(1.718) \) gives us a value less than \( 1 \), confirming that \( c \) lies within the interval \((0, 1)\).

### Conclusion
The value of \( c \) that satisfies Lagrange's Mean Value Theorem for the function \( f(x) = e^x \) in the interval \([0, 1]\) is:

\[ c = \ln(e - 1) \]