Question

Please help! Will mark the brainliest!

Answer 1

The domain is the set of all possible x-values which will make the function valid.
$f(x) = \frac{6}{x+3} \ \ \ \ , \ g(x) = \frac{1}{4-x}$
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) $f + g = \frac{6}{x+3} + \frac{1}{4-x} = \frac{6(4-x)+(x+3)}{(x+3)(4-x)}$
The domain of (f+g) ⇒⇒⇒ R - {-3,4}
because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4


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


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

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

Because ⇒⇒⇒  x+3 = 0  ⇒⇒⇒ x =-3

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

because: x+3 = 0 ⇒⇒⇒ x = -3   and    4 - x = 0 ⇒⇒⇒ x = 4
(8) $\frac{g}{f} = \frac{ \frac{1}{4-x} }{ \frac{6}{x+3} } = \frac{x+3}{6(4-x)}$
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) $(f+g)(x) = \frac{6}{x+3} + \frac{1}{4-x} = \frac{6(4-x)+(x+3)}{(x+3)(4-x)}$

∴ $(f + g)(x) = \frac{24-6x+x+3}{(x+3)(4-x)} = \frac{27-5x}{(x+3)(4-x)}$

(10) $(f - g)(x) = \frac{6}{x+3} - \frac{1}{4-x} = \frac{6(4-x)-(x+3)}{(x+3)(4-x)}$

∴ $(f - g)(x) = \frac{24-6x-x-3}{(x+3)(4-x)} = \frac{21 - 7x}{(x+3)(4-x)}$

(11) $(f * g)(x) = \frac{6}{x+3} * \frac{1}{4-x} = \frac{6}{(x+3)(4-x)}$


(12) $(f * f)(x) = \frac{6}{x+3} * \frac{6}{x+3} = \frac{36}{(x+3)^2}$


(13) $(\frac{f}{g})(x) = \frac{\frac{6}{x+3} }{ \frac{1}{4-x} } = \frac{6(4-x)}{x+3}$


(14) $(\frac{g}{f})(x) = \frac{ \frac{1}{4-x} }{ \frac{6}{x+3} } = \frac{x+3}{6(4-x)}$

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