Match up each function with its domain and range. They form a set of domino cards that can be arranged in a sequence.
To turn the sequence into a closed loop, you will need to complete the cards with missing information. A separate sheet containing just those cards is available here .
\(f(x)=x^2+1\)
domain:\(x\neq-1\) \(f(x)\neq0\)
\(f(x)=|x|\)
domain:\(x\in \mathbb{R}\) \(f(x)\ge1\)
\(f(x)=\sqrt[3]{x}\)
domain:\(x\in \mathbb{R}\) \(0\!<\!f(x)\!\le\!1\)
\(f(x)=\sqrt{x}\)
domain:\(x\in \mathbb{R}\) \(f(x)\ge-1\)
\(f(x)=\frac{1}{x}\)
domain:\(x\neq0\) \(f(x)\ne-1\)
\(f(x)=x+\frac{1}{x}\)
domain:\(x\neq-1\) , \(x\neq1\) \(f(x)\le-1\) or \(f(x)>0\)
\(f(x)=\frac{1}{x^2}\)
domain:\(x\in \mathbb{R}\) \(f(x)\ge0\)
\(f(x)=\frac{1}{x^2+1}\)
domain:\(x\ge0\) \(f(x)\ge0\)
\(f(x)=\frac{1}{x^2-1}\)
domain:\(x>0\) \(f(x)>0\)
\(f(x)=\frac{1}{1+x}\)
domain:\(x\ge0\) \(f(x)\le\frac{2}{3}\)
\(f(x)=\frac{1}{\sqrt{x}}\)
domain:\(x\in \mathbb{R}\) \(f(x)\in \mathbb{R}\)
\(f(x)=\sqrt{3x}(1-x)\)
domain:\(x\neq0\) \(f(x)\neq0\)
\(f(x)=\frac{1}{x}-1\)
domain:
\(f(x)=\qquad\qquad\)
domain:
