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Given: \( f(x)=\frac{1}{x+5} \) and \( g(x)=x-2 \). What are the restrictions of the domain of \( f(g(x)) \) ? \( x=-5 \) \( x \neq-3 \) \( 0 x \neq 2 \) 0 There are no restrictions. DONE \( \mathbf{x} \)

Ask by Burgess French. in the United States
Mar 21,2025

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Tutor-Verified Answer

Answer

The domain of \( f(g(x)) \) is all real numbers except \( x = -3 \).

Solution

1. Substitute \( g(x) \) into \( f(x) \) to obtain the composite function: \[ f(g(x)) = f(x-2) = \frac{1}{(x-2)+5}. \] 2. Simplify the expression: \[ (x-2)+5 = x+3, \] so \[ f(g(x)) = \frac{1}{x+3}. \] 3. For the function \( f(g(x)) \) to be defined, its denominator must be nonzero. Set the denominator not equal to zero: \[ x+3 \neq 0. \] 4. Solve for \( x \): \[ x \neq -3. \] Thus, the restriction on the domain of \( f(g(x)) \) is \( x \neq -3 \).

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To find the restrictions of the domain for \( f(g(x)) \), you need to evaluate the function composition. First, substitute \( g(x) = x - 2 \) into \( f(x) \), giving you \( f(g(x)) = f(x - 2) = \frac{1}{(x-2)+5} = \frac{1}{x+3} \). The restriction arises from the denominator: it cannot equal zero. This leads to \( x + 3 \neq 0 \) or \( x \neq -3 \). Therefore, the only restriction is \( x \neq -3 \). Another point to consider is that \( g(x) \) is defined for all \( x \) since it's just a linear function. Thus, the restriction on \( f(g(x)) \) solely comes from the behavior of the function \( f(x) \). Therefore, \( x \neq -3 \) is the singular focus for the domain restriction of the composition of those functions.

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