If `f (x) = sin x+cos x, g(x)= x^2 - 1,` then gif (x)) is invertible in domain

Options

A. `[(-pi)/(2) ,0]`
B. ` [-(pi)/(4),(pi)/(4)]`
C. `[-(pi)/(2),pi]`
D. `(0,pi)`

Answer

To determine if the function \( g(f(x)) \) is invertible in a given domain, we need to follow these steps:

### Step 1: Define the functions
We have:
- \( f(x) = \sin x + \cos x \)
- \( g(x) = x^2 - 1 \)

### Step 2: Find the composition \( g(f(x)) \)
To find \( g(f(x)) \), we substitute \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g(\sin x + \cos x) = (\sin x + \cos x)^2 - 1 \]

### Step 3: Simplify \( g(f(x)) \)
Now, we simplify \( g(f(x)) \):
\[ g(f(x)) = (\sin x + \cos x)^2 - 1 \]
Using the identity \( (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x \) and knowing that \( \sin^2 x + \cos^2 x = 1 \):
\[ g(f(x)) = 1 + 2\sin x \cos x - 1 = 2\sin x \cos x \]
Using the double angle identity, we can express this as:
\[ g(f(x)) = \sin(2x) \]

### Step 4: Determine the intervals for invertibility
To check if \( g(f(x)) = \sin(2x) \) is invertible, we need to find intervals where it is one-to-one. The sine function is one-to-one in the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) for \( 2x \), which corresponds to \( x \) in the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \).

### Step 5: Conclusion
Thus, \( g(f(x)) \) is invertible in the interval \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \).