Question

One question please help!

Determine the domain of (f/g)(x)when f(x)=-1/x and g(x)=(sqrt)3x-9

Answer 1

The domain of (f/g)(x) will be:   {$x\in R : x>3$}

Explanation

Given functions are:  $f(x)= -\frac{1}{x}$ and $g(x)=\sqrt{3x-9}$

So...

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

As $(f/g)(x)$ is in fraction form, so the denominator part can't be 0 for defining the function and as (3x-9) is under square root so the value of (3x-9) will be always greater than 0

Thus.....

$x\neq 0\\ \\ or \\ \\ 3x-9>0\\ 3x>9\\ x>3$

So the domain of (f/g)(x) will be:  {$x\in R : x>3$}

Answer 2

Question 1: Determine the domain of h(x)=$\frac{5}{\sqrt{2x+10} }$ and write it using set builder notation.

Answer: {x|x>-5}

Question 2: Find the domain and range of the function shown in the graph. Write the domain and range using interval notation.

Answer: domain: (-6, -1] ∪ (1, 5]

              range: (-3, 2] ∪ [3, 7)

Question 3: Find (f*g)(x) when f(x)=$\frac{\sqrt{x+3} }{x}$ and g(x)= $\frac{\sqrt{x+3} }{2x}$

Answer: $\frac{x+3}{2x^{2} }$

Question 4: Determine the domain of (f/g)(x) when f(x)=$\frac{-1}{x}$ and g(x)=$\sqrt{3x-9}$

Answer: (3,∞)

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