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enter image description here This is my working/how I've answered the question. However, does anyone know how I would state the domain of each function using set notation?

My working:

For g(x), the domain is when the inside of the square root is greater than or equal to zero. Thus, x² - 1 ≥ 0 x² ≥ 1 Upon taking the square root of both sides, we result in: x ≥ 1 or x ≤ -1 (the principle square root gets the positive, and the secondary square root gets the negative with the inequality sign flipped).

Thus, this function g(x) has the domain:

{x such that -1 ≥ x ≥ 1 } Or alternatively, {x such that |x| ≥ 1}

With regards to f(x), the domain of the cosine function are all real numbers. The domain of a composite function is the domain which satisfies each individual function.

Thus, the domain of the composite function is: {x such that |x| ≥ 1}

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    $\begingroup$ It is correct, now do the b)! $\endgroup$ – user90628 Aug 20 '13 at 8:42
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If $f,~~g$ are functions such that $f\circ g$ be a function so

$$\text{dom}(f\circ g)\subseteq \text{dom}(g)$$ and $$\text{dom}(f\circ g)=\text{dom}(g)\ \Longleftrightarrow \text{rang}(g)\subseteq \text{dom}(f)$$

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  • $\begingroup$ I will give him +1 so that he have 33,333 $\endgroup$ – user90628 Aug 20 '13 at 8:50
  • $\begingroup$ Now, Congratulations 33,333 reputation with 3 kinds of badges including 3 gold one:D $\endgroup$ – Mahdi Khosravi Aug 20 '13 at 8:53
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    $\begingroup$ @Mahdi Khosrav And his answer has 3 upvotes (at the moment). $\endgroup$ – user90628 Aug 20 '13 at 8:56
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    $\begingroup$ And all numbers of his badges are divisible by 3. $\endgroup$ – user90628 Aug 20 '13 at 8:58
  • $\begingroup$ +1 & Lots of supportive comments for you dear friend :^) $\endgroup$ – Namaste Aug 21 '13 at 0:27

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