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  1. $f: R \rightarrow R$ where $f = \{(x, \sqrt{x})\mid x \in R \}$
  2. $f: R \rightarrow R$ where $f = \{(x, \tan{x})\mid x \in R\}.$

I believe the $1^{\text{st}}$ one would be considered a function despite negative $R$ values not giving real roots. We can just define the domain to be non-negative.

For the second, in case of some domain values, $f(x)$ would exist but would be infinite. Therefore, it would be considered a function too despite not having a well-defined set of $(x, f(x)).$

I would like someone to verify if my approach is correct.

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    $\begingroup$ Your functions are ill-defined because the target space is not right. Your function as defined take values in $R\times R$. (What you actually define as $f$ are the graphs of the functions $x \mapsto \sqrt{x}$ and $x \mapsto \tan(x)$. $\endgroup$ Sep 28, 2020 at 9:07
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    $\begingroup$ For 1) as you says you have to consider $x \in \mathbb R^+$ $\endgroup$ Sep 28, 2020 at 9:19
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    $\begingroup$ For 2) you have to consider that for argument $\dfrac {\pi}{2}$ (and its multiples) the value is not a number $\endgroup$ Sep 28, 2020 at 9:20

1 Answer 1

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A function must be defined on every element of its domain, meaning the image of each element of the domain must lie in the codomain. Assuming $R$ is meant to be the set $\mathbb{R}$ of all real numbers, that is not the case here.

Since the square root of a negative number is an imaginary number, $f(-1)$ is not a real number (never mind the issue of how to define what $\sqrt{-1}$ means). Hence, we cannot define a function $f: \mathbb{R} \to \mathbb{R}$ by $f = \{(x, \sqrt{x}) \mid x \in \mathbb{R}\}$. To ensure that $f$ were a function, we would have to restrict the domain to $[0, \infty)$.

The tangent of a real number $x$ is defined by $$\tan x = \frac{\sin x}{\cos x}$$ for each $x \in \mathbb{R}$ such that $\cos x \neq 0$. However, $\cos x = 0$ whenever $$x = \frac{\pi}{2} + n\pi, n \in \mathbb{Z}$$ Thus, $f: \mathbb{R} \to \mathbb{R}$ defined by $f = \{(x, \tan x) \mid x \in \mathbb{R}\}$ is not a function since it is not defined at those values of $x$ where $\cos x = 0$. In this case, to define a function, we would have to restrict the domain to those values of $x \in \mathbb{R}$ where $\cos x \neq 0$.

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