Let `f : R ^(+) to R` defined as `f (x)= e ^(x) + `ln x and `g = f ^(-1)` then correct statement (s) is/are:

Options

A. `g''(e)=(1-e)/((1+e)^(3))`
B. `g ''(e) (e-1)/((1+e)^(3))`
C. `g '(e) =e+1`
D. `g '(e) =(1)/(e+1)`

Answer

To solve the problem, we need to analyze the function \( f(x) = e^x + \ln x \) and its inverse \( g = f^{-1} \). We will find the first and second derivatives of \( g \) at a specific point, which is \( e \).

### Step 1: Find \( g(e) \)
Since \( g \) is the inverse of \( f \), we need to find \( g(e) \) by solving \( f(x) = e \).

\[ f(x) = e^x + \ln x = e \]

We can check for \( x = 1 \):

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

Thus, \( g(e) = 1 \).

### Step 2: Find \( g'(x) \)
Using the formula for the derivative of the inverse function, we have:

\[ g'(x) = \frac{1}{f'(g(x))} \]

Substituting \( x = e \):

\[ g'(e) = \frac{1}{f'(g(e))} = \frac{1}{f'(1)} \]

### Step 3: Find \( f'(x) \)
Now we calculate \( f'(x) \):

\[ f'(x) = \frac{d}{dx}(e^x + \ln x) = e^x + \frac{1}{x} \]

Evaluating at \( x = 1 \):

\[ f'(1) = e^1 + \frac{1}{1} = e + 1 \]

Thus,

\[ g'(e) = \frac{1}{e + 1} \]

### Step 4: Find \( g''(x) \)
Using the formula for the second derivative of the inverse function:

\[ g''(x) = -\frac{f''(g(x))}{(f'(g(x)))^3} \]

Substituting \( x = e \):

\[ g''(e) = -\frac{f''(g(e))}{(f'(g(e)))^3} = -\frac{f''(1)}{(f'(1))^3} \]

### Step 5: Find \( f''(x) \)
Now we calculate \( f''(x) \):

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

Evaluating at \( x = 1 \):

\[ f''(1) = e^1 - \frac{1}{1^2} = e - 1 \]

### Step 6: Calculate \( g''(e) \)
Now we can substitute \( f''(1) \) and \( f'(1) \) into the equation for \( g''(e) \):

\[ g''(e) = -\frac{e - 1}{(e + 1)^3} \]

### Conclusion
The correct statements regarding \( g'(e) \) and \( g''(e) \) can be derived from the above calculations.