Question

Given that \( f(x) = \sin x + \cos x \) and \( g(x) = x^2 - 1 \), find the conditions under which \( g(f(x)) \) is invertible.

• A. \( -\frac{\pi}{4} \leq x \leq \frac{\pi}{4} \)
• B. \( 0 \leq x \leq \pi \)
• C. \( -\frac{\pi}{4} \leq x \leq \pi \)
• D. \( 0 \leq x \leq \frac{\pi}{2} \)

Hint:

Invertibility requires the function to be monotonic (either strictly increasing or strictly decreasing) over the interval.

Answer 1

The Correct Option is A

Step 1: Find \( g[f(x)] \)
We are given \( f(x) = \sin x + \cos x \) and \( g(x) = x^2 - 1 \). We need to find \( g[f(x)] \), which means \( g(f(x)) \). \[ g[f(x)] = g(\sin x + \cos x) \] Substitute \( \sin x + \cos x \) in place of \( x \) in the expression for \( g(x) \): \[ g[f(x)] = (\sin x + \cos x)^2 - 1 \] Step 2: Expand and simplify \( g[f(x)] \)
Expand the square: \[ g[f(x)] = (\sin^2 x + 2\sin x \cos x + \cos^2 x) - 1 \] Recall the trigonometric identities: \[ \sin^2 x + \cos^2 x = 1 \] \[ 2\sin x \cos x = \sin 2x \] Substitute these identities into the expression: \[ g[f(x)] = (1 + \sin 2x) - 1 \] \[ g[f(x)] = \sin 2x \] Step 3: Analyze the monotonicity of \( \sin 2x \)
The function \( \sin 2x \) is a sinusoidal function. We need to find an interval where it is strictly increasing or strictly decreasing (i.e., monotonic) to ensure invertibility. 
Step 4: Determine an interval for invertibility
The sine function is strictly increasing in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). Since we have \( \sin 2x \), we need to consider the argument \( 2x \). For \( \sin 2x \) to be strictly increasing, we need \( 2x \) to be in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). \[ -\frac{\pi}{2} \le 2x \le \frac{\pi}{2} \] Divide by 2: \[ -\frac{\pi}{4} \le x \le \frac{\pi}{4} \] In this interval \( [-\frac{\pi}{4}, \frac{\pi}{4}] \), \( \sin 2x \) is strictly increasing, and therefore invertible. 
Conclusion: \( g[f(x)] = \sin 2x \). The function \( g[f(x)] \) is invertible in the interval \( [-\frac{\pi}{4}, \frac{\pi}{4}] \).