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Consider the following two functions:

$f(x) = \sqrt {4 - x^2};\;\{x : x \in \mathbb{R}, x \ge -2 \,\ \text{and} \ \, x \le 2\}$

$g(x) = \sqrt {1 + x};\;\{x : x \in \mathbb{R}, x \ge -1\}$

Given that:

$$\frac{f(x)}{g(x)} = \frac{\sqrt{4-x^2}}{\sqrt{1+x}} = \sqrt{\frac{4-x^2}{1+x}}$$

Is the domain of $\frac{f(x)}{g(x)}$ determined by the first step or only the last?

For instance, does the domain include $-3$?

If the domain is determined by the first step, then the answer would be NO, because both $\sqrt{4-(-3)^2}$ and $\sqrt{1+(-3)}$ are imaginary.

However, if the domain is determined by the second step, then the answer would be YES, because $\sqrt{\frac{4-(-3)^2}{1+(-3)}} = \sqrt{\frac{-5}{-2}} = \sqrt{\frac{5}{2}}$

Any help appreciated.

Robert Howard
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jwool
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    This is a good question: at its core, it boils down to how you define a function. A function is not defined by an expression, but by a domain and and an expression (let's say, for simplicity; that second part is not entirely accurate). But the curriculum seems to forget that last part and only define it via an expression, which is the source of quite a lot of confusion... – Clement C. Feb 25 '19 at 21:59
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    Might be worth noting that $\frac {\sqrt{4 - x^2}}{\sqrt{1+x}} \ne \sqrt{\frac {4-x^2}{1+x}}$ which makes the question moot. – fleablood Feb 25 '19 at 22:02
  • @fleablood Then ask the same question with $f=g$ defined on $\mathbb{R}\setminus{0}$ by $f(x)=1/x$, with the question being whether $0$ is in the domain? – Clement C. Feb 25 '19 at 22:04
  • Maybe, but we still have the question of pole removing, for instance $f(x)=(x-x_0)h(x)$ and $g(x)=(x-x_0)$ or even continuity extension like $\dfrac{\sin(x)}x$. – zwim Feb 25 '19 at 22:05
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    Great first question @jwool. – Thomas Davis Feb 26 '19 at 00:52

2 Answers2

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This only depends on your definition of your new function (call it $h(x)$).

If $h : x \mapsto \frac{f(x)}{g(x)}$ then the domain of $h$ is limited to the intersection of domains of $g$ and $f$.

If you define $h : x \mapsto \sqrt{\frac{4-x^2}{1+x}}$ then the domain of $h$ can be larger

Thomas Lesgourgues
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Interesting question. As you have it set up, the answer is no.

Let $h(x)=\frac{f(x)}{g(x)}$, and assume $x$ was in the domain of $h(x)$. Then $h(x)$ exists, which means by definition that $f(x)$ and $g(x)$ exist. So $x$ is in the domain of $f$ and $g$

NazimJ
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