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Find functions \( f \) and \( g \) so that \( f \circ g=H \). H(x) \( f 6 x+6 \mid \) Choose the correct pair of functions. A. \( f(x)=\frac{x-6}{6}, g(x)=|-x| \) B. \( f(x)=|x|, g(x)=6 x+6 \) C. \( f(x)=|-x|, g(x)=\frac{x-6}{6} \) D. \( f(x)=6 x+6, g(x)=|x| \)

Ask by Henry Clarke. in the United States
Mar 06,2025

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There is no correct pair of functions among the options that satisfy \( f \circ g = H \).

Solución

To find functions \( f \) and \( g \) such that \( f \circ g = H \), we need to understand what \( H(x) = 6x + 6 \) means in terms of function composition. The composition \( f(g(x)) \) should equal \( 6x + 6 \). Let's analyze each option step by step: ### Option A: - \( f(x) = \frac{x - 6}{6} \) - \( g(x) = |-x| \) Calculating \( f(g(x)) \): \[ g(x) = |-x| \implies f(g(x)) = f(|-x|) = \frac{|-x| - 6}{6} \] This does not simplify to \( 6x + 6 \). ### Option B: - \( f(x) = |x| \) - \( g(x) = 6x + 6 \) Calculating \( f(g(x)) \): \[ g(x) = 6x + 6 \implies f(g(x)) = f(6x + 6) = |6x + 6| \] This does not simplify to \( 6x + 6 \). ### Option C: - \( f(x) = |-x| \) - \( g(x) = \frac{x - 6}{6} \) Calculating \( f(g(x)) \): \[ g(x) = \frac{x - 6}{6} \implies f(g(x)) = f\left(\frac{x - 6}{6}\right) = \left| -\frac{x - 6}{6} \right| = \frac{6 - x}{6} \] This does not simplify to \( 6x + 6 \). ### Option D: - \( f(x) = 6x + 6 \) - \( g(x) = |x| \) Calculating \( f(g(x)) \): \[ g(x) = |x| \implies f(g(x)) = f(|x|) = 6|x| + 6 \] This does not simplify to \( 6x + 6 \). ### Conclusion: None of the options provided yield \( f(g(x)) = 6x + 6 \). Therefore, it appears that there is no correct pair of functions among the given options that satisfy \( f \circ g = H \). If you have any further questions or need additional assistance, feel free to ask!

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Beyond the Answer

To find the correct pair of functions \( f \) and \( g \) such that \( f \circ g = H(x) = 6x + 6 \), we need to first express what \( f(g(x)) \) should equal. Let's analyze each option: A. \( f(x)=\frac{x-6}{6}, g(x)=|-x| \) - Here, \( g(x) = |-x| \) makes \( g(x) \) always non-negative. Plugging this into \( f \), we get: \[ f(g(x)) = f(|-x|) = \frac{|-x| - 6}{6} \] This does not simplify to \( 6x + 6 \), so A is incorrect. B. \( f(x)=|x|, g(x)=6x+6 \) - Plugging in \( g(x) \): \[ f(g(x)) = f(6x + 6) = |6x + 6| \] This cannot simplify to \( 6x + 6 \) for negative \( x \), so B is incorrect. C. \( f(x)=|-x|, g(x)=\frac{x-6}{6} \) - Plugging in \( g(x) \): \[ f(g(x)) = f\left(\frac{x-6}{6}\right) = \left|-\frac{x-6}{6}\right| = \frac{x-6}{6} \] This does not equal \( 6x + 6 \), so C is incorrect. D. \( f(x)=6x+6, g(x)=|x| \) - Plugging in \( g(x) \): \[ f(g(x)) = f(|x|) = 6|x| + 6 \] Again, this does not simplify directly to \( 6x + 6 \) since it depends on the sign of \( x \). After going through each option, it appears none of the pairs correctly result in \( f \circ g = H(x) = 6x + 6 \). However, Option D seems to be a plausible candidate if we are only looking for non-negative \( x \) inputs. So, while strictly none of the options works for all \( x \), D would be the most reasonable choice for positive inputs, effectively outputting \( H(x) \) when \( x \ge 0 \).

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