The domain is the set of all possible
x-values which will make the function valid.
[tex]f(x) = \frac{3}{x-2} \ \ \ \ , \ g(x) = \sqrt{x-1} [/tex]
For the given function :
The denominator of a fraction cannot be
zero
The number under a square root sign must be
Non negative
(a)
(1) The domain of f ⇒⇒⇒ R - {2}
Because ⇒⇒⇒ x - 2 = 0 ⇒⇒⇒ x = 2
(2) The domain of g ⇒⇒⇒ [1,∞)
Because: x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1
(3) [tex]f + g = \frac{3}{x-2} + \sqrt{x-1} [/tex]
The domain of (f+g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2 and x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1
(4) [tex]f - g = \frac{3}{x-2} - \sqrt{x-1}[/tex]
The domain of (f-g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2 and x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1
(5) [tex]f * g = \frac{3}{x-2} * \sqrt{x-1} = \frac{3 \sqrt{x-1}}{(x-2)} [/tex]
The domain of (f*g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2 and x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1
(6) [tex]f * f = \frac{3}{x-2} * \frac{3}{x-2} = \frac{9}{(x-2)^2}[/tex]
The domain of ff ⇒⇒⇒ R - {2}
Because ⇒⇒⇒ x-2 = 0 ⇒⇒⇒ x = 2
(7) [tex] \frac{f}{g} = \frac{\frac{3}{x-2} }{ \sqrt{x-1} } = \frac{3}{(x-2) \sqrt{x-1}} [/tex]
The domain of (f/g) ⇒⇒⇒ (1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2 and x - 1 > 0 ⇒⇒⇒ x > 1
(8) [tex] \frac{g}{f} = \frac{ \sqrt{x-1} }{ \frac{3}{x-2} } = \frac{1}{3} (x-2) \sqrt{x-1} [/tex]
The domain of (g/f) ⇒⇒⇒ [1,∞) - {2}
Because: x - 2 = 0 ⇒⇒⇒ x = 2 and x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1
===================================================
(b)
(9) [tex] (f + g)(x) = \frac{3}{x-2} + \sqrt{x-1} [/tex]
(10) [tex](f - g)(x) = \frac{3}{x-2} - \sqrt{x-1}[/tex]
(11) [tex](f * g)(x) = \frac{3}{x-2} * \sqrt{x-1} = \frac{3 \sqrt{x-1}}{(x-2)} [/tex]
(12) [tex](f * f)(x) = \frac{3}{x-2} * \frac{3}{x-2} = \frac{9}{(x-2)^2}[/tex]
(13) [tex] \frac{f}{g} = \frac{\frac{3}{x-2} }{ \sqrt{x-1} } = \frac{3}{(x-2) \sqrt{x-1}} [/tex]
(14) [tex] (\frac{g}{f})(x) = \frac{ \sqrt{x-1} }{ \frac{3}{x-2} } = \frac{1}{3} (x-2) \sqrt{x-1} [/tex]
