Answer 1

Given:

\[f\left( x \right) = \sqrt{x + 1}\text{ and } g\left( x \right) = \sqrt{9 - x^2}\]

Clearly,

\[f\left( x \right) = \sqrt{x + 1}\]  is defined for all x ≥ - 1.
Thus, domain (f) = [1, ∞]
Again,

 

\[g\left( x \right) = \sqrt{9 - x^2}\]   is defined for  9 -x² ≥ 0 ⇒ x² - 9 ≤ 0

⇒ x² - 3² ≤ 0

⇒ (x + 3)(x - 3) ≤ 0
⇒ \[x \in \left[ - 3, 3 \right]\]

Thus, domain (g) = [ - 3, 3]

Now,

domain ( f ) ∩ domain( g ) = [ -1, ∞] ∩ [- 3, 3]    = [ -1, 3]

(iv) \[\frac{f}{g}: \left[ - 1, 3 \right] \to \text{ R is given by } \left( \frac{f}{g} \right)\left( x \right) = \frac{f\left( x \right)}{g\left( x \right)} = \frac{\sqrt{x + 1}}{\sqrt{9 - x^2}} = \sqrt{\frac{x + 1}{9 - x^2}}\].