If `f(x)=sinx+cosx `and `g(x)=x^2-1`, then `g(f (x)) `is invertible in the domain .

Options

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

Explanation

To determine if the function \( g(f(x)) \) is invertible, we need to analyze the functions \( f(x) \) and \( g(x) \) given by:

- \( f(x) = \sin x + \cos x \)
- \( g(x) = x^2 - 1 \)

### Step 1: Find the composition \( g(f(x)) \)

First, we need to compute \( g(f(x)) \):

\[ g(f(x)) = g(\sin x + \cos x) = (\sin x + \cos x)^2 - 1 \]

### Step 2: Expand \( g(f(x)) \)

Now we expand \( (\sin x + \cos x)^2 \):

\[ (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x \]

Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):

\[ g(f(x)) = 1 + 2\sin x \cos x - 1 = 2\sin x \cos x \]

### Step 3: Simplify \( g(f(x)) \)

We can simplify \( 2\sin x \cos x \) using the double angle identity:

\[ g(f(x)) = \sin(2x) \]

### Step 4: Determine the invertibility of \( g(f(x)) \)

A function is invertible if it is one-to-one (1-1) and onto (onto).

#### Checking if \( g(f(x)) = \sin(2x) \) is one-to-one:

The function \( \sin(2x) \) is periodic with a period of \( \pi \). Therefore, it is not one-to-one over its entire domain. However, we can restrict the domain to make it one-to-one.

To ensure that \( \sin(2x) \) is one-to-one, we can restrict \( x \) to the interval where \( 2x \) spans from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). This corresponds to:

\[ x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \]

In this interval, \( \sin(2x) \) is strictly increasing, thus it is one-to-one.

### Step 5: Determine if \( g(f(x)) \) is onto

The range of \( \sin(2x) \) as \( x \) varies in \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \) is from \( -1 \) to \( 1 \). Therefore, it covers the entire codomain of \( \sin(2x) \) which is also \( [-1, 1] \).

### Conclusion

Since \( g(f(x)) = \sin(2x) \) is one-to-one and onto in the restricted domain \( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \), we conclude that \( g(f(x)) \) is invertible in this domain.

### Final Answer

The function \( g(f(x)) \) is invertible in the domain \( x \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \).

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