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 \( [-\frac{\pi}{4}, \frac{\pi}{4}] \). 
 

Answer 2

Step 1: Analyze the function \( g(f(x)) \)
We are given two functions:
\( f(x) = \sin x + \cos x \) and \( g(x) = x^2 - 1 \).
We need to determine the conditions under which the composite function \( g(f(x)) \) is invertible.

Step 2: Express the composite function
The composite function \( g(f(x)) \) is defined as:
\[ g(f(x)) = g(\sin x + \cos x) = (\sin x + \cos x)^2 - 1. \] Expanding the square: \[ g(f(x)) = (\sin x + \cos x)^2 - 1 = \sin^2 x + 2 \sin x \cos x + \cos^2 x - 1. \] Using the identity \( \sin^2 x + \cos^2 x = 1 \), we simplify: \[ g(f(x)) = 1 + 2 \sin x \cos x - 1 = 2 \sin x \cos x. \] Using the double angle identity for sine, \( 2 \sin x \cos x = \sin(2x) \), we have: \[ g(f(x)) = \sin(2x). \] Thus, the composite function is: \[ g(f(x)) = \sin(2x). \]

Step 3: Determine the conditions for invertibility
For a function to be invertible, it must be one-to-one (i.e., it must be either strictly increasing or strictly decreasing). The function \( \sin(2x) \) is one-to-one on an interval where it does not repeat its values.
The function \( \sin(2x) \) is one-to-one in the interval where the sine function does not repeat its values. This occurs when the argument \( 2x \) lies within \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), because the sine function is one-to-one in this range.
So, we need to find the interval for \( x \) such that \( 2x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).
Dividing by 2, we get: \[ -\frac{\pi}{4} \leq x \leq \frac{\pi}{4}. \]

Step 4: Conclusion
Thus, the function \( g(f(x)) \) is invertible when \( x \) lies in the interval \( \left[ -\frac{\pi}{4}, \frac{\pi}{4} \right] \).
The correct answer is:

\( -\frac{\pi}{4} \leq x \leq \frac{\pi}{4} \)