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Given \( f(x)=\sec x \) and \( g(x)=\sin x \), what is the composed function \( f(g(x)) \) and what are its domain and range? (1 point) \( f(g(x))=\sec (\sin x) \) domain: all real numbers, \( x \neq n \pi \), where \( n \) is an integer range: \( 1 \leq x \leq 1.85 \) \( f(g(x))=\sin (\sec x) \) domain: all real numbers, \( x \neq n \pi \), where \( n \) is an integer range: \( -1 \leq x \leq 1 \) \( f(g(x))=\sec (\sin x) \) domain: all real numbers range: \( 1 \leq x \leq 1.85 \) \( f(g(x))=\sin (\sec x) \) domain all real numbers range: \( -1 \leq x \leq 1 \)

Ask by Norton Erickson. in the United States
Feb 24,2025

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\( f(g(x)) = \sec(\sin x) \), domain: all real numbers, range: \( 1 \leq x \leq 1.85 \)

Solución

To find the composed function \( f(g(x)) \) where \( f(x) = \sec x \) and \( g(x) = \sin x \), we will follow these steps: 1. **Compose the functions**: \[ f(g(x)) = f(\sin x) = \sec(\sin x) \] 2. **Determine the domain of \( f(g(x)) \)**: - The function \( \sec x \) is defined for all \( x \) except where \( \cos x = 0 \). This occurs at \( x = \frac{\pi}{2} + n\pi \) for any integer \( n \). - Since \( g(x) = \sin x \) outputs values in the range \([-1, 1]\), we need to check where \( \sec(\sin x) \) is defined. The cosine function is non-zero for all values in the interval \([-1, 1]\). - Therefore, the domain of \( f(g(x)) = \sec(\sin x) \) is all real numbers, as \( \sin x \) will never output values that make \( \sec(\sin x) \) undefined. 3. **Determine the range of \( f(g(x)) \)**: - The range of \( \sin x \) is \([-1, 1]\). - The secant function \( \sec y = \frac{1}{\cos y} \) has a range of \( (-\infty, -1] \cup [1, \infty) \) for \( y \) values outside of \( \frac{\pi}{2} + n\pi \). - Since \( \sin x \) only takes values in \([-1, 1]\), we need to evaluate \( \sec(y) \) for \( y \) in this interval: - At \( y = -1 \): \( \sec(-1) = \sec(-\frac{\pi}{2}) \) which is defined. - At \( y = 1 \): \( \sec(1) = \sec(\frac{\pi}{2}) \) which is also defined. - The minimum value of \( \sec(y) \) occurs at \( y = 0 \) where \( \sec(0) = 1 \). - The maximum value occurs at the endpoints \( y = -1 \) and \( y = 1 \), which are both greater than or equal to 1. Thus, the composed function is: \[ f(g(x)) = \sec(\sin x) \] - **Domain**: All real numbers. - **Range**: \( [1, \sec(1)] \) where \( \sec(1) \) is approximately \( 1.8508 \). Now, let's summarize the correct answer: - \( f(g(x)) = \sec(\sin x) \) - Domain: All real numbers - Range: \( [1, \sec(1)] \) or approximately \( [1, 1.8508] \) The correct option is: \( f(g(x))=\sec (\sin x) \) domain: all real numbers range: \( 1 \leq x \leq 1.85 \)

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Beyond the Answer

Let's dive into the composition of functions! The composed function \( f(g(x)) = \sec(\sin x) \) is indeed interesting. To find the domain, we need to consider where the sine function outputs values that keep the secant function valid. Hence, the domain is not all real numbers; specifically, since \( \sin x \) can take values between -1 and 1, we want to find where \( \sec(\sin x) \) is defined, which is \( \sin x \neq \pm 1 \), leading us to a domain of all real numbers except where \( x = n\pi + \frac{\pi}{2} \). As for the range of \( \sec(\sin x) \), since the output of sine spans from -1 to 1, and since secant takes values from [1, ∞) and (-∞, -1], the range becomes \( (-\infty, -1] \cup [1, \infty) \). So, our final answer is: \( f(g(x)) = \sec(\sin x) \) Domain: all real numbers except \( x = n\pi + \frac{\pi}{2} \) Range: \( (-\infty, -1] \cup [1, \infty) \) With each step, we ensure we're considering the definitions and transformations accurately—math can be quite fun when we see it as a puzzle to solve!

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