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Chapter 1: Problem 80

Show that $$(g \circ f)(x)=x \text { and }(f \circ g)(x)=x$$ $$f(x)=x^{3}-1, g(x)=\sqrt[3]{x+1}$$

Short Answer

Expert verified
Both \( (g \circ f)(x) = x\) and \( (f \circ g)(x) = x\), therefore both compositions result in the identity function.

Step by step solution

01

Calculating (g 鈼� f)(x)

First calculate the composition of \(g\) and \(f\), or \(g(f(x))\). Start by replacing \(x\) in \(f(x)\) with \(x^{3} -1\), resulting in: \(g(f(x)) = g(x^{3} -1)\). Then replace \(x\) in function \(g\) with the function \(f(x) = x^{3} -1\), resulting in: \(g(f(x)) = \sqrt[3]{(x^{3} -1)+1} = \sqrt[3]{x^{3}} = x\).
02

Calculating (f 鈼� g)(x)

Now calculate the composition of \(f\) and \(g\), or \(f(g(x))\). Start by replacing \(x\) in \(g(x)\) with \(\sqrt[3]{x+1}\), resulting in: \(f(g(x)) = f(\sqrt[3]{x+1})\). Then replace \(x\) in function \(f\) with the function \(g(x) = \sqrt[3]{x+1}\), resulting in: \(f(g(x)) = (\sqrt[3]{x+1})^{3} -1 = x +1 -1 = x\).

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Most popular questions from this chapter

The notation \(\left.f(x)\right|_{a} ^{b}\) is used to denote the difference \(f(b)-f(a) .\) That is, $$\left.f(x)\right|_{a} ^{b}=f(b)-f(a)$$ Evaluate \(\left.f(x)\right|_{0} ^{b}\) for the given function \(f\) and the indicated values of \(a\) and \(b\). $$f(x)=2 x^{3}-3 x^{2}-x ;\left.f(x)\right|_{0} ^{2}$$

If \(y=x^{2}-3 x+2,\) find \(x\) when \(y=0 .[1.1]\)

Solve by completing the square or by using the quadratic formula. $$x^{2}-3 x+5$$

A fixed point of a function is a number \(a\) such that \(f(a)=a\) . Find all fixed points for the given function. $$g(x)=\frac{x}{x+5}$$

Use a graphing utility. Graph \(f(x)=\frac{[[2 x]]}{|x|}\) for \(-4 \leq x \leq 4\) and \(x \neq 0\)
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