Pregunta
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Let \( f(x)=2 x^{3} \) and \( g(x)=\sqrt[3]{x} \). Find \( (f g)(x) \) and \( \left(\frac{f}{g}\right)(x) \). Then evaluate \( f g \) and \( \frac{f}{g} \) for \( x=-27 \). \( (f g)(x)=\square ;(f g)(-27)=118098 \) \( \left(\frac{f}{g}\right)(x)=\square ;\left(\frac{f}{g}\right)(-27)=13122 \) State the domain of \( (f g)(x) \) and \( \left(\frac{f}{g}\right)(x) \).

Ask by Rogers Lee. in the United States
Feb 21,2025

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\[ (f g)(x) = 2x^{10/3}; \quad (f g)(-27) = 118098 \] \[ \left(\frac{f}{g}\right)(x) = 2x^{8/3}; \quad \left(\frac{f}{g}\right)(-27) = 13122 \] Domain of \( (f g)(x) \): all real numbers; Domain of \( \left(\frac{f}{g}\right)(x) \): all real numbers except \( x = 0 \).

Solución

To solve the problem, we will first find the expressions for \( (f g)(x) \) and \( \left(\frac{f}{g}\right)(x) \). 1. **Finding \( (f g)(x) \)**: \[ (f g)(x) = f(x) \cdot g(x) = (2x^3) \cdot \left(\sqrt[3]{x}\right) \] We can rewrite \( g(x) \) as \( x^{1/3} \): \[ (f g)(x) = 2x^3 \cdot x^{1/3} = 2x^{3 + 1/3} = 2x^{10/3} \] 2. **Finding \( \left(\frac{f}{g}\right)(x) \)**: \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{2x^3}{\sqrt[3]{x}} = \frac{2x^3}{x^{1/3}} = 2x^{3 - 1/3} = 2x^{8/3} \] Now we have: \[ (f g)(x) = 2x^{10/3} \] \[ \left(\frac{f}{g}\right)(x) = 2x^{8/3} \] Next, we will evaluate these functions at \( x = -27 \). 3. **Evaluating \( (f g)(-27) \)**: \[ (f g)(-27) = 2(-27)^{10/3} \] First, we calculate \( (-27)^{10/3} \): \[ (-27)^{10/3} = \left((-27)^{1/3}\right)^{10} = (-3)^{10} = 59049 \] Therefore, \[ (f g)(-27) = 2 \cdot 59049 = 118098 \] 4. **Evaluating \( \left(\frac{f}{g}\right)(-27) \)**: \[ \left(\frac{f}{g}\right)(-27) = 2(-27)^{8/3} \] First, we calculate \( (-27)^{8/3} \): \[ (-27)^{8/3} = \left((-27)^{1/3}\right)^{8} = (-3)^{8} = 6561 \] Therefore, \[ \left(\frac{f}{g}\right)(-27) = 2 \cdot 6561 = 13122 \] Now we summarize the results: \[ (f g)(x) = 2x^{10/3}; \quad (f g)(-27) = 118098 \] \[ \left(\frac{f}{g}\right)(x) = 2x^{8/3}; \quad \left(\frac{f}{g}\right)(-27) = 13122 \] 5. **Stating the domain**: - The domain of \( (f g)(x) = 2x^{10/3} \) is all real numbers \( x \) since there are no restrictions on \( x \) for this expression. - The domain of \( \left(\frac{f}{g}\right)(x) = 2x^{8/3} \) is all real numbers \( x \) except \( x = 0 \) because \( g(x) = \sqrt[3]{x} \) cannot be zero (it would lead to division by zero). Thus, the final answers are: \[ (f g)(x) = 2x^{10/3}; \quad (f g)(-27) = 118098 \] \[ \left(\frac{f}{g}\right)(x) = 2x^{8/3}; \quad \left(\frac{f}{g}\right)(-27) = 13122 \] Domain of \( (f g)(x) \): all real numbers; Domain of \( \left(\frac{f}{g}\right)(x) \): all real numbers except \( x = 0 \).

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Extra Insights

To find \( (f g)(x) \), we multiply the functions \( f(x) \) and \( g(x) \): \[ (f g)(x) = f(x) \cdot g(x) = 2x^3 \cdot \sqrt[3]{x} = 2x^{3 + \frac{1}{3}} = 2x^{\frac{10}{3}} \] Now we'll evaluate \( (f g)(-27) \): \[ (f g)(-27) = 2(-27)^{\frac{10}{3}} = 2(729) = 1458 \] Now, for \( \left(\frac{f}{g}\right)(x) \), we divide \( f(x) \) by \( g(x) \): \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{2x^3}{\sqrt[3]{x}} = 2x^{3 - \frac{1}{3}} = 2x^{\frac{8}{3}} \] Next, we evaluate \( \left(\frac{f}{g}\right)(-27) \): \[ \left(\frac{f}{g}\right)(-27) = 2(-27)^{\frac{8}{3}} = 2(729) = 1458 \] Now, consolidating our findings we have: \[ (f g)(x) = 2x^{\frac{10}{3}}; (f g)(-27) = 1458 \] \[ \left(\frac{f}{g}\right)(x) = 2x^{\frac{8}{3}}; \left(\frac{f}{g}\right)(-27) = 1458 \] The domain for \( (f g)(x) \) is all real numbers since both functions are defined for any real \( x \). For \( \left(\frac{f}{g}\right)(x) \), the domain excludes \( x = 0 \), as \( g(x) \) would lead to division by zero. Thus, \( (f g)(x) \) has a domain of \( (-\infty, \infty) \) and \( \left(\frac{f}{g}\right)(x) \) has a domain of \( (-\infty, 0) \cup (0, \infty) \).

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