Let `f(x) =1+2sin(e^x/(e^x+1))` `x ge 0` then `f^(-1)(x) ` is equal to (assuming f is bijectve)

Options

A. `log ((sin^(-1)((x-1)/2))/(1 - sin^(-1) (( x-1)/2))) `
B. `log ((sin (( x-1)/2))/(1- sin((x-1)/2)))`
C. `e^((sin^(-1).((x-1)/2))/(1-sin^(-1)((x-1)/2)))`
D. `e^((sin((x-1)/2))/(1-sin ((x-1)/2)))`

Answer

To solve the problem, we need to find the inverse function \( f^{-1}(x) \) for the function given by

\[ f(x) = 1 + 2 \sin\left(\frac{e^x}{e^x + 1}\right) \]

where \( x \geq 0 \).

### Step 1: Set \( f(x) = y \)

We start by setting

\[ y = f(x) = 1 + 2 \sin\left(\frac{e^x}{e^x + 1}\right) \]

### Step 2: Rearrange the equation

Rearranging the equation gives us:

\[ y - 1 = 2 \sin\left(\frac{e^x}{e^x + 1}\right) \]

### Step 3: Isolate the sine function

Now, we isolate the sine function:

\[ \sin\left(\frac{e^x}{e^x + 1}\right) = \frac{y - 1}{2} \]

### Step 4: Apply the inverse sine function

Next, we apply the inverse sine function:

\[ \frac{e^x}{e^x + 1} = \sin^{-1}\left(\frac{y - 1}{2}\right) \]

### Step 5: Rearrange to isolate \( e^x \)

Now, we rearrange this equation to isolate \( e^x \):

\[ e^x = (e^x + 1) \sin^{-1}\left(\frac{y - 1}{2}\right) \]

This leads to:

\[ e^x - e^x \sin^{-1}\left(\frac{y - 1}{2}\right) = \sin^{-1}\left(\frac{y - 1}{2}\right) \]

### Step 6: Factor out \( e^x \)

Factoring out \( e^x \) gives us:

\[ e^x (1 - \sin^{-1}\left(\frac{y - 1}{2}\right)) = \sin^{-1}\left(\frac{y - 1}{2}\right) \]

### Step 7: Solve for \( e^x \)

Now, we can solve for \( e^x \):

\[ e^x = \frac{\sin^{-1}\left(\frac{y - 1}{2}\right)}{1 - \sin^{-1}\left(\frac{y - 1}{2}\right)} \]

### Step 8: Take the natural logarithm

Taking the natural logarithm of both sides gives us:

\[ x = \log\left(\frac{\sin^{-1}\left(\frac{y - 1}{2}\right)}{1 - \sin^{-1}\left(\frac{y - 1}{2}\right)}\right) \]

### Step 9: Substitute back to find \( f^{-1}(x) \)

Now, substituting \( y \) back with \( x \) (since we are looking for \( f^{-1}(x) \)), we have:

\[ f^{-1}(x) = \log\left(\frac{\sin^{-1}\left(\frac{x - 1}{2}\right)}{1 - \sin^{-1}\left(\frac{x - 1}{2}\right)}\right) \]

### Final Result

Thus, the inverse function is:

\[ f^{-1}(x) = \log\left(\frac{\sin^{-1}\left(\frac{x - 1}{2}\right)}{1 - \sin^{-1}\left(\frac{x - 1}{2}\right)}\right) \]