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.
$f(x) = \frac{3}{x-2} \ \ \ \ , \ g(x) = \sqrt{x-1}$
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) $f + g = \frac{3}{x-2} + \sqrt{x-1}$
The domain of (f+g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1


(4) $f - g = \frac{3}{x-2} - \sqrt{x-1}$
The domain of (f-g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1


(5) $f * g = \frac{3}{x-2} * \sqrt{x-1} = \frac{3 \sqrt{x-1}}{(x-2)}$
The domain of (f*g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1

(6) $f * f = \frac{3}{x-2} * \frac{3}{x-2} = \frac{9}{(x-2)^2}$
The domain of ff ⇒⇒⇒ R - {2}

Because ⇒⇒⇒  x-2 = 0  ⇒⇒⇒ x = 2

(7) $\frac{f}{g} = \frac{\frac{3}{x-2} }{ \sqrt{x-1} } = \frac{3}{(x-2) \sqrt{x-1}}$
The domain of (f/g) ⇒⇒⇒ (1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 > 0 ⇒⇒⇒ x > 1

(8) $\frac{g}{f} = \frac{ \sqrt{x-1} }{ \frac{3}{x-2} } = \frac{1}{3} (x-2) \sqrt{x-1}$
The domain of (g/f) ⇒⇒⇒ [1,∞) - {2}
Because: x - 2 = 0 ⇒⇒⇒ x = 2   and   x - 1 ≥ 0  ⇒⇒⇒ x ≥ 1
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(b)


(9) $(f + g)(x) = \frac{3}{x-2} + \sqrt{x-1}$


(10) $(f - g)(x) = \frac{3}{x-2} - \sqrt{x-1}$


(11) $(f * g)(x) = \frac{3}{x-2} * \sqrt{x-1} = \frac{3 \sqrt{x-1}}{(x-2)}$


(12) $(f * f)(x) = \frac{3}{x-2} * \frac{3}{x-2} = \frac{9}{(x-2)^2}$


(13) $\frac{f}{g} = \frac{\frac{3}{x-2} }{ \sqrt{x-1} } = \frac{3}{(x-2) \sqrt{x-1}}$


(14) $(\frac{g}{f})(x) = \frac{ \sqrt{x-1} }{ \frac{3}{x-2} } = \frac{1}{3} (x-2) \sqrt{x-1}$