Malone Carrillo
06/02/2024 · Junior High School

or the real-valued functions \( f(x)=x^{2}-1 \) and \( g(x)=\sqrt{x-6} \), find the composition \( f \circ g \) and specify its domain using hterval notation.

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To find the composition \( f \circ g \), we need to substitute \( g(x) \) into \( f(x) \). So, we have: \[ f(g(x)) = f(\sqrt{x-6}) \] Now, we substitute \( \sqrt{x-6} \) for \( x \) in \( f(x) \): \[ f(\sqrt{x-6}) = (\sqrt{x-6})^2 - 1 \] Squaring the square root will eliminate the square root: \[ f(\sqrt{x-6}) = x - 6 - 1 \] Simplify the expression: \[ f(\sqrt{x-6}) = x - 7 \] Now, we need to find the domain of \( f \circ g \). The domain of \( f \circ g \) is the set of all \( x \) values for which both \( g(x) \) and \( f(x) \) are defined. The domain of \( g(x) = \sqrt{x-6} \) is all \( x \) such that \( x-6 \geq 0 \), because the square root function is only defined for non-negative arguments. So, the domain of \( g(x) \) is: \[ x \geq 6 \] Since \( f(x) = x^2 - 1 \) is a polynomial function, it is defined for all real numbers. Therefore, the domain of \( f \circ g \) is the same as the domain of \( g(x) \), which is: \[ x \in [6, \infty) \] So, the composition \( f \circ g \) is \( x - 7 \), and its domain is \( [6, \infty) \).

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The composition \( f \circ g \) is \( x - 7 \), and its domain is \( [6, \infty) \).
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