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The textbook seeks the domain of $f/g$, where $ f(x)= \sqrt {x }$ and $g(x) = |x-3|$. The answer stated is $(0,\infty]$. I have two questions here:

  1. Shouldn't $3$ be excluded from the domain as one can't divide by zero?
  2. Is it okay to use square brackets for infinity? I have never seen infinity included in the domain of any function before, and my teacher clearly stated that infinity is always followed by a parenthesis, not a square bracket. The answer to a similar problem seeking the domain of $g(x) = |x-3|$ is (-$\infty$,$\infty$]. How? Please explain this as well.
Ambica Govind
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You are correct on both counts, and are missing one other count. Because $f$ is defined at $0$, and $g$ is defined and nonzero at $0$, $\frac fg$ is also defined at $0$.

The expression $\frac{\sqrt{x}}{|x-3|}$ is defined for values of $x\in[0,3)\cup(3,\infty)$. Your textbook sounds weird and might be using some nonstandard notation, but without knowing the textbook, I can't be the judge of that.

5xum
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  • It's the 14th edition of Thomas' Calculus. – Ambica Govind Nov 22 '21 at 08:08
  • @AmbicaGovind I don't have access to the book, so all I can say is that the correct answer is $[0,3)\cup(3,\infty)$. I don't know why the answer in the book is $(0,\infty]$, but in commonly used mathematical notation, that answer is just plain wrong. – 5xum Nov 22 '21 at 08:11
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    The only reasonable explanation is that the autor mixed $f$ and $g$. In fact $\frac{|x-3|}{\sqrt{x}}$ is defined on $(0,\infty)$. – Paul Frost Nov 22 '21 at 10:06
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    @PaulFrost Even that doesn't really explain the closed brackets at infinity though... – 5xum Nov 22 '21 at 10:12