I'm given a function: $$f\left(x\right)=\left(\frac{1}{x-4}\right)$$ I have to find it's domain and prove it. Finding the domains is the easy part $(-\infty,4)\cup(4,\infty)$. But how do i prove it?
How to prove the domain of a function?
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calculus
algebra-precalculus
functions
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0Let $D:=\mathbb{R}\backslash\{4\}$ and $x\in D$. Then $x-4\neq0$ so $(x-4)^{-1}$ exists in $\mathbb{R}$. Assuming $\operatorname{dom}(f)$ is to be as large as possible then we have $D\subseteq\operatorname{dom}(f)$. Also $4\not\in\operatorname{dom}(f)$ because in that case $x-4=0$ has no multiplicative inverse in $\mathbb{R}$. So $\operatorname{dom}(f)\subseteq D$. All in all, $\operatorname{dom}(f)=D$. – 2017-01-26
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0@NeedForHelp you should leave this as an answer – 2017-01-26
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2You can't have a function without already having the domain. What you have is an expression. Given an expression, we can then ask what is the largest subset of $\mathbb R$ or $\mathbb C$ for which the expression makes sense. This set is often called the "natural domain" associated with the expression. – 2017-01-26
1 Answers
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Let $D:=\mathbb{R}\backslash\{4\}$ and $x\in D$. Then $x-4\neq0$ so $(x-4)^{-1}$ exists in $\mathbb{R}$. Assuming $\operatorname{dom}(f)$ is to be as large as possible then we have $D\subseteq\operatorname{dom}(f)$. Also $4\not\in\operatorname{dom}(f)$ because in that case $x-4=0$ has no multiplicative inverse in $\mathbb{R}$. So $\operatorname{dom}(f)\subseteq D$. All in all, $\operatorname{dom}(f)=D$.