Text Let f(x)=x+1f(x)=\sqrt{x+1}f(x)=x+1 and g(x)=x2−1g(x)=x^{2}-1g(x)=x2−1. (a) Simplify (f∘g)(x)(f\circ g)(x)(f∘g)(x) and find its domain. (b) Compute (f∘g)(−5)(f\circ g)(-5)(f∘g)(−5). Solution (a) (f∘g)(x)=g(x)+1=(x2−1)+1=x2=∣x∣(f\circ g)(x)=\sqrt{g(x)+1}=\sqrt{(x^{2}-1)+1}=\sqrt{x^{2}}=|x|(f∘g)(x)=g(x)+1=(x2−1)+1=x2=∣x∣. The radicand x2≥0x^{2}\ge 0x2≥0 for every xxx, so D=RD=\mathbb{R}D=R. (b) (f∘g)(−5)=∣−5∣=5(f\circ g)(-5)=|-5|=5(f∘g)(−5)=∣−5∣=5. (f∘g)(x)=∣x∣, D=R,(f∘g)(−5)=5 \boxed{\,(f\circ g)(x)=|x|,\ D=\mathbb{R},\quad (f\circ g)(-5)=5\,}(f∘g)(x)=∣x∣, D=R,(f∘g)(−5)=5