Additional Mathematics – Functions | e-Consult

Answer 3

1. \((p\circ q)(x)=p(q(x))=p(x^{2}-5)=\sqrt{(x^{2}-5)+4}=\sqrt{x^{2}-1}\).
   

   For the square‑root to be defined, we need \(x^{2}-1\ge 0\) ⇒ \(|x|\ge 1\).
   

   Domain of \((p\circ q)\): \((-\infty,-1]\cup[1,\infty)\).

2. \((q\circ p)(x)=q(p(x))=q(\sqrt{x+4})=(\sqrt{x+4})^{2}-5 = (x+4)-5 = x-1\).
   

   The inner function \(p(x)=\sqrt{x+4}\) requires \(x+4\ge0\) ⇒ \(x\ge-4\).
   

   Thus the composition is defined for \(x\ge-4\).
   

   Domain of \((q\circ p)\): \([ -4,\infty )\).

3. \((p\circ q)(x)=\sqrt{x^{2}-1}\) is not one‑one over its whole domain because \(\sqrt{x^{2}-1}\) yields the same value for \(x\) and \(-x\) (e.g., \(x=2\) and \(x=-2\) both give \(\sqrt{3}\)).
   

   If we restrict the domain to \([1,\infty)\) (or alternatively to \((-\infty,-1]\)), the function becomes one‑one and therefore invertible.
   

   Taking the restriction \(x\ge1\):
   

   \(y=\sqrt{x^{2}-1}\) ⇒ \(y^{2}=x^{2}-1\) ⇒ \(x^{2}=y^{2}+1\) ⇒ \(x=\sqrt{y^{2}+1}\) (positive root because \(x\ge1\)).
   

   Hence the inverse on this restricted domain is \((p\circ q)^{-1}(y)=\sqrt{y^{2}+1}\) with domain \(y\ge0\) (since square‑root outputs are non‑negative) and range \([1,\infty)\).