Respuesta :

The domain is the set of all possible x-values which will make the function valid.
[tex]f(x) = \frac{6}{x+3} \ \ \ \ , \ g(x) = \frac{1}{4-x} [/tex]
For the given function The denominator of a fraction cannot be zero

(a)

(1) The domain of f ⇒⇒⇒ R - {-3}

Because ⇒⇒⇒  x+3 = 0  ⇒⇒⇒ x =-3

(2) The domain of g ⇒⇒⇒ R - {4}
Because: 4 - x = 0 ⇒⇒⇒ x = 4

(3) [tex]f + g = \frac{6}{x+3} + \frac{1}{4-x} = \frac{6(4-x)+(x+3)}{(x+3)(4-x)} [/tex]
The domain of (f+g) ⇒⇒⇒ R - {-3,4}
because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4


(4) [tex]f - g = \frac{6}{x+3} - \frac{1}{4-x} = \frac{6(4-x)-(x+3)}{(x+3)(4-x)} [/tex]
The domain of (f-g) ⇒⇒⇒ R - {-3,4}
because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4


(5) [tex]f * g = \frac{6}{x+3} * \frac{1}{4-x} = \frac{6}{(x+3)(4-x)} [/tex]
The domain of (f*g) ⇒⇒⇒ R - {-3,4}
because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4

(6) [tex]f * f = \frac{6}{x+3} * \frac{6}{x+3} = \frac{36}{(x+3)^2}[/tex]
The domain of ff ⇒⇒⇒ R - {-3}

Because ⇒⇒⇒  x+3 = 0  ⇒⇒⇒ x =-3

(7) [tex] \frac{f}{g} = \frac{\frac{6}{x+3} }{ \frac{1}{4-x} } = \frac{6(4-x)}{x+3} [/tex]
The domain of (f/g) ⇒⇒⇒ R - {-3,4}

because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4
(8) [tex] \frac{g}{f} = \frac{ \frac{1}{4-x} }{ \frac{6}{x+3} } = \frac{x+3}{6(4-x)} [/tex]
The domain of (g/f) ⇒⇒⇒ R - {-3,4}

because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4
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(b)


(9) [tex] (f+g)(x) = \frac{6}{x+3} + \frac{1}{4-x} = \frac{6(4-x)+(x+3)}{(x+3)(4-x)} [/tex]

∴ [tex](f + g)(x) = \frac{24-6x+x+3}{(x+3)(4-x)} = \frac{27-5x}{(x+3)(4-x)} [/tex]

(10) [tex](f - g)(x) = \frac{6}{x+3} - \frac{1}{4-x} = \frac{6(4-x)-(x+3)}{(x+3)(4-x)} [/tex]

∴ [tex](f - g)(x) = \frac{24-6x-x-3}{(x+3)(4-x)} = \frac{21 - 7x}{(x+3)(4-x)} [/tex]

(11) [tex](f * g)(x) = \frac{6}{x+3} * \frac{1}{4-x} = \frac{6}{(x+3)(4-x)} [/tex]


(12) [tex](f * f)(x) = \frac{6}{x+3} * \frac{6}{x+3} = \frac{36}{(x+3)^2}[/tex]


(13) [tex] (\frac{f}{g})(x) = \frac{\frac{6}{x+3} }{ \frac{1}{4-x} } = \frac{6(4-x)}{x+3} [/tex]


(14) [tex] (\frac{g}{f})(x) = \frac{ \frac{1}{4-x} }{ \frac{6}{x+3} } = \frac{x+3}{6(4-x)} [/tex]

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