If \( f(x)=\ln x, g(x)=e^{3 x} \), and \( h(x)=x^{4} \), find the following. (a) \( (f \circ g)(x) \). What is the domain of \( f \circ g \) ? (b) \( (g \circ f)(x) \). What is the domain of \( g \circ f \) ? (c) \( (f \circ g)(5) \) (d) \( (f \circ h)(x) \). What is the domain of \( f \circ h \) ? (e) \( (f \circ h)(e) \)
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Let's solve each of the functions one by one. (a) The composition \( (f \circ g)(x) \) means \( f(g(x)) = f(e^{3x}) \). Since \( f(x) = \ln x \), we have: \[ (f \circ g)(x) = \ln(e^{3x}) = 3x. \] The domain of \( f \circ g \) is determined by the domain of \( g(x) \), since \( g(x) \) must yield a valid input for \( f(x) \). The function \( g(x) = e^{3x} \) is defined for all real numbers, and since \( e^{3x} > 0 \) for all \( x \), the domain of \( f \circ g \) is all real numbers \( (-\infty, \infty) \). (b) The composition \( (g \circ f)(x) \) means \( g(f(x)) = g(\ln x) \). So, we have: \[ (g \circ f)(x) = e^{3 \ln x} = x^{3}. \] The domain of \( g \circ f \) is dictated by the domain of \( f(x) = \ln x \), which is \( (0, \infty) \) because the logarithm is only defined for positive values of \( x \). (c) To find \( (f \circ g)(5) \): \[ (f \circ g)(5) = 3 \cdot 5 = 15. \] (d) For \( (f \circ h)(x) \), we compute \( f(h(x)) = f(x^4) \): \[ (f \circ h)(x) = \ln(x^4) = 4 \ln x. \] The domain of \( f \circ h \) is also constrained by \( f(x) \), requiring \( x^4 > 0 \). Therefore, the domain is \( x > 0 \) or \( (0, \infty) \). (e) To evaluate \( (f \circ h)(e) \): \[ (f \circ h)(e) = 4 \ln(e) = 4 \cdot 1 = 4. \] So, the final results are: - \( (f \circ g)(x) = 3x \) with domain \( (-\infty, \infty) \). - \( (g \circ f)(x) = x^3 \) with domain \( (0, \infty) \). - \( (f \circ g)(5) = 15 \). - \( (f \circ h)(x) = 4 \ln x \) with domain \( (0, \infty) \). - \( (f \circ h)(e) = 4 \).