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Chapter 1: Problem 37

Find (a) \(f \circ g\) and (b) \(g \circ f .\) Find the domain of each function and each composite function. $$f(x)=\sqrt[3]{x-5}, \quad g(x)=x^{3}+1$$

Short Answer

Expert verified
The compositions are \(f(g(x)) = \sqrt[3]{x^{3}-4}\) and \(g(f(x)) = x - 4\). The domains of \(f(x)\),\(g(x)\), and \(f(g(x))\) are all real numbers, while the domain of \(g(f(x))\) is \(x \geq 5\).

Step by step solution

01

Compose Function f(g(x))

Compose the functions \(f(x)\) and \(g(x)\) to find \(f(g(x))\). This involves substituting \(x^{3}+1\) (which is \(g(x)\)) into the formula of \(f(x)\). The composition \(f \circ g = f(g(x))\) would be \( \sqrt[3]{(x^{3}+1)-5}\). Simplify this to obtain the function \(f \circ g(x) = \sqrt[3]{x^{3}-4}\).
02

Compose Function g(f(x))

Next we find \(g \circ f(x)\), which means substituting \(\sqrt[3]{x-5}\) (f(x)) into \(g(x)\). Therefore, \(g \circ f = g(f(x))\) would be \((\sqrt[3]{x-5})^{3}+1\). Simplifying this equation gives \(g \circ f(x) = x - 4\).
03

Calculate the domains

The domain of a function is the set of all possible values of the input x that make the function defined. The original functions \(f(x) = \sqrt[3]{x-5}\) and \(g(x) = x^{3}+1\) are defined for all real numbers. The composite function \(f(g(x)) = \sqrt[3]{x^{3} - 4}\) is also defined for all real numbers. Similarly, the composite function \(g(f(x)) = x - 4\) is defined for all \(x \geq 5\). The reason being for the last function, the cube root function won't be defined for \(x < 5\).

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