Question 3.5.4: Let f and g be real functions defined by f(x) = √x - 1 and g......

Clearly, f(x)=\sqrt{x-1} is defined for all real values of x for which x-1 \geq 0 , i.e., x \geq 1 . So, \operatorname{dom}(f)=[1, \infty) .

Also, g(x)=\sqrt{x+1} is defined for all real values of x for which x+1 \geq 0 , i.e., x \geq-1 . So, \operatorname{dom}(g)=[-1, \infty) .

\therefore \quad \operatorname{dom}(f) \cap \operatorname{dom}(g)=\{1, \infty) \cap[-1, \infty)=[1, \infty) .

(i) (f+g):[1, \infty) \rightarrow R is given by

(ii) (f-g):[1, \infty) \rightarrow R is given by

(iii) (f g):[1, \infty) \rightarrow R is given by