Math  /  Algebra

QuestionIf f(x)=ab(xc)2+df(x)=\frac{a}{b}(x-c)^{2}+d, and g(x)=f1(x)g(x)=f^{-1}(x), a restriction to the domain of f(x)f(x) that will ensure g(x)g(x) is a function is {xxc,xR}\{x \mid x \geq c, x \in \mathbb{R}\}
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Confirm that the restriction allows g(x)=f1(x) g(x) = f^{-1}(x) to be a function.
Since f(x) f(x) is one-to-one on the domain xc x \geq c , it has an inverse function g(x)=f1(x) g(x) = f^{-1}(x) . This inverse will map each y y value back to exactly one x x value, satisfying the definition of a function.
The restriction {xxc,xR}\{x \mid x \geq c, x \in \mathbb{R}\} ensures that g(x)=f1(x) g(x) = f^{-1}(x) is a function.

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