Identifying Where Functions Are Continuous

Determine where each function is continuous:

1. \(F(x) =\sqrt{x^{2}+4}\)
2. \(G(x) =\sqrt{x^{2}-1}\)
3. \(H(x) =\dfrac{x^{2}-1}{x^{2}-4}+\sqrt{x-1}\)

Solution (a) \(F=f\circ g\) is the composite of \(f( x) =\sqrt{x}\) and \(g( x) =x^{2}+4\). \(f\) is continuous for \(x\geq 0\) and \(g\) is continuous for all real numbers. The domain of \(F\) is all real numbers and \(F=( f\circ g) ( x) =\sqrt{x^{2}+4}\) is continuous for all real numbers. That is, \(F\) is continuous on its domain.

(b) \(G\) is the composite of \(f( x) =\sqrt{x}\) and \( g( x) =x^{2}-1\). \(f\) is continuous for \(x\geq 0\) and \(g\) is continuous for all real numbers. The domain of \(G\) is \(\{ x|x\geq 1\} \cup \{ x|x\leq -1\} \) and \(G=( f\circ g) ( x) =\sqrt{x^{2}-1}\) is continuous on its domain.

(c) \(H\) is the sum of \(f( x) =\) \(\dfrac{x^{2}-1}{ x^{2}-4}\) and the function \(g( x) =\sqrt{x-1}\). The domain of \(f\) is \(\{ x|x\neq -2, x\neq 2\} \); \(f\) is continuous on its domain. The domain of \(g\) is \(x\geq 1\). The domain of \(H\) is \(\{ x|1\leq x\lt 2\} \cup \{ x|x>2\}\); \(H\) is continuous on its domain.