Domains of f(x)=1/sqrt(x) and f'

ahorn

ahorn

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Hello



Is the domain of \(\displaystyle f(x)=\dfrac{1}{\sqrt{x}}\) equal to \(\displaystyle x\geq 0\)? I am assuming this because x is the expression within a square root so cannot be non-negative.


And is the domain of \(\displaystyle f'(x)=-\dfrac{1}{2x\sqrt{x}}\) equal to \(\displaystyle x > 0 \)? I am assuming that \(\displaystyle x \neq 0 \) here because I am assuming that f is not differentiable on its endpoint.
 
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Yes. Your domains are correct. The derivative has x in the denominator, so x cannot be 0. The concern is -'s in the square roots (or other even roots such as 4th roots) and 0's in the denominator. You can verify answers for problems like this at wolframalpha.com.
 
Ok thanks. I'll try use wolframalpha next time. From what you are saying about the denominator not equalling 0, doesn't that mean that the domain of f also has \(\displaystyle x \neq 0\)?
 
Whoops. From the post timestamps, it looks like ahorn edited his post while fcabanski was typing.

The domain statement for f is not currently correct because 0 is not in the domain.

:)
 
ahorn said:
Is the domain of \(\displaystyle f(x)=\dfrac{1}{\sqrt{x}}\) equal to \(\displaystyle x\geq 0\)? I am assuming this because x is the expression within a square root so cannot be non-negative.
Click to expand...

The phrase highlighted in red above is not correct. Is that a typo?

You need to fix the inequality symbol, in the domain statement above.

:cool:

Edit: After precalculus, one ought to be able to reason through the domain of 1/sqrt(x), without resorting to technology.
 
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Both domains are x>0. I didn't see that the OP wrote >= in the first case.

"Resorting to technology" is fine to check an answer.

Too many students refuse to use a calculator, or aren't taught to use it, to help with problem solving. For example, if you graph 1/sqrt(x) you'll see the domain clearly.
 
Quaid said:
That is a conditional statement, my friend. :D




Are you talkin' problem-solving or answer-checking?
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I am talking about both. It's valid to graph a function to see its domain. In this problem, for example, the first thing I'd do is graph it. Then I can see the domain not only to arrive at the answer, but to verify the answer.

"Resorting to technology is fine" is a conditional statement. The condition is being in the real world.
 
fcabanski said:
The condition is being in the real world.
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Is there some other kind of world, in which we can be, lol?

Yes, yes, technology is a fine thing (I use it every day), yet, I believe from the perspective of learning that too much of a fine thing ain't always a good thing. We need reasoning skills, for the real world (power failures, system failures, et al).

By the way, what happens when the domain is (sqrt[101]/7,sqrt[201]/7)? How do you get those values from a graph?

:idea: The real world is not always an ideal world, but mental effort saves the day (we hope).
 
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