Coiliposition of two functions: Advanced For the real-valued functions \( f(x)=\sqrt{2 x+12} \) and \( g(x)=x-3 \), find the composition \( f \circ g \) and specify its \( d \) \[ (f \circ g)(x)= \] Domain of \( f \circ g \) : \( \square \)

To find the composition of the functions \( f(x) = \sqrt{2x + 12} \) and \( g(x) = x - 3 \), we need to substitute the expression for \( g(x) \) into \( f(x) \) and simplify. Given: \( f(x) = \sqrt{2x + 12} \) \( g(x) = x - 3 \) The composition \( f \circ g \) is found by substituting \( g(x) \) into \( f(x) \): \( (f \circ g)(x) = f(g(x)) = f(x - 3) = \sqrt{2(x - 3) + 12} \) Simplifying the expression: \( (f \circ g)(x) = \sqrt{2x - 6 + 12} = \sqrt{2x + 6} \) Therefore, the composition \( f \circ g \) is \( \sqrt{2x + 6} \). To find the domain of \( f \circ g \), we need to ensure that the expression under the square root is non-negative. The expression under the square root is \( 2x + 6 \), which must be greater than or equal to 0. Solving the inequality \( 2x + 6 \geq 0 \): \( 2x \geq -6 \) \( x \geq -3 \) Therefore, the domain of \( f \circ g \) is \( x \geq -3 \).