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This question must be weird, but:

Is $f(x) = 1$ equal to $f(x) = \frac{x}{x}$ ?

The reason I am asking is:

Since $\frac{x}{x} = 1$, it would be reasonable to conclude $f(x) = 1$ is indeed equal to $f(x) = \frac{x}{x}$. However, it we look at the graphs of these two functions, we can see that $f(x) = 1$ is continuous everywhere, yet $f(x) = \frac{x}{x}$ is undefined at zero.

Stokolos Ilya
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3 Answers3

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It is $$\frac{x}{x}=1$$ only if $$x\neq 0$$

Dr. Sonnhard Graubner
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Actually, this has nothing to do with continuity. You defined your functions through analytic expressions, with no mention to their domains. However, assuming that the domain of each of them is the set of all real numbers for which the corresponding analytic expression makes sense, then they have different domains and therefore they cannot be the same function.

José Carlos Santos
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  • They could have the same domains however, they haven't been specified and so it is ambiguous whether by $f(x)=1$ we are referring to $f~:~\Bbb R\to \Bbb R$ or if we are referring to $\Bbb R^*\to\Bbb R$ or if we are referring to ${1}\to{1}$, etc... The function $f(x)=1$ could be bijective depending on choice of domain and codomain. – JMoravitz Aug 12 '19 at 11:52
  • It is worth remembering that when poorly phrased textbooks ask "find the domain of the function" they really are asking "find the maximal subset of $\Bbb R$ which can act as a domain for the function" instead, but there is no reason to assume that this is the only valid domain. – JMoravitz Aug 12 '19 at 11:54
  • @JMoravitz I agree and I've edited my answer. – José Carlos Santos Aug 12 '19 at 12:12
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Obviously not! The reason is the one you mention! $f(x)=1$ is defined on $\mathbb{R}$ whereas $f(x)=\frac{x}{x}$ is defined on $\mathbb{R} \setminus \{ 0 \}$.

Note: Before you simplify a function take care of the domain first.

Tolaso
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