Question

Find f + g, f − g, cf (c ∈ R, c ≠ 0), fg, \[\frac{1}{f}\text{  and } \frac{f}{g}\] in : 

(b) If \[f\left( x \right) = \sqrt{x - 1}\]  and  \[g\left( x \right) = \sqrt{x + 1}\]

 

Answer 1

Given: \[f\left( x \right) = \sqrt{x - 1}\] and \[g\left( x \right) = \sqrt{x + 1}\] Thus,
(f + g) ) : [1, ∞) → R is defined by (f + g) (x) = f (x) + g (x) = \[\sqrt{x - 1} + \sqrt{x + 1}\] (f -  g) ) : [1, ∞) → R is defined by (f - g) (x) = f (x) -  g (x) = \[\sqrt{x - 1} - \sqrt{x + 1}\] cf : [1, ∞) → R is defined by (cf) (x) = \[c\sqrt{x - 1}\] (fg) : [1, ∞) → R is defined by (fg) (x) = f(x).g(x) = (fg) :

[1, ∞) → R is defined by (fg) (x) = f(x).g(x) = \[\sqrt{x - 1} \times \sqrt{x + 1} = \sqrt{x^2 - 1}\]

\[\frac{1}{f}: \left( 1, \infty \right) \to \text{ R isdefined by } \left( \frac{1}{f} \right)\left( x \right) = \frac{1}{f\left( x \right)} = \frac{1}{\sqrt{x - 1}} . \] \[\frac{f}{g}: [1, \infty ) \to \text{ R is defined by } \left( \frac{f}{g} \right)\left( x \right) = \frac{f\left( x \right)}{g\left( x \right)} = \frac{\sqrt{x - 1}}{\sqrt{x + 1}} = \sqrt{\frac{x - 1}{x + 1}} .\]