For the real-valued functions $f(x)=x^{2}-1$ and $g(x)=\sqrt{x+4}$, find the composition $f \circ g$ and specify its domain using interval notation
\[
\begin{array}{l}
\qquad(f \circ g)(x)=\square \\
\text { Domain of } f \circ g: \square
\end{array}
\]
Answer
\(\boxed{\text{Domain of } f \circ g: [-4, \infty)}\)
Step 1 ：\(f(g(x)) = (\sqrt{x+4})^2 - 1\)
Step 2 ：\(f(g(x)) = x + 4 - 1 = x + 3\)
Step 3 ：\(g(x) = \sqrt{x+4}\) is defined for all \(x\) such that \(x+4 \geq 0\), so \(x \geq -4\)
Step 4 ：\(f(g(x)) = x + 3\) is defined for all real numbers. However, since \(g(x)\) is part of the composition, we must also consider the domain of \(g(x)\). Therefore, the domain of \(f \circ g\) is all \(x\) such that \(x \geq -4\)
Step 5 ：\(\boxed{(f \circ g)(x) = x + 3}\)
Step 6 ：\(\boxed{\text{Domain of } f \circ g: [-4, \infty)}\)
