Consider y = −√x. Is y a function of x? Why or why not? Is it true that |y| is always less than x? Show that it is or give a counter example.

clairestdean

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Consider y = −√x. Is y a function of x? Why or why not? Is it true that |y| is always less than x? Show that it is or give a counter example.
 

Subhotosh Khan

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Consider y = −√x. Is y a function of x? Why or why not? Is it true that |y| is always less than x? Show that it is or give a counter example.
Please share your work/thoughts about this assignment.

Please follow the rules of posting in this forum, as enunciated at:

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clairestdean

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I think that y is a function of x because graphing the function I get f(x)=−√x has the same solutions as y = −√x.. I'm not sure if this makes sense though.
 

Subhotosh Khan

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I think that y is a function of x because graphing the function I get f(x)=−√x has the same solutions as y = −√x.. I'm not sure if this makes sense though.
The graph of a function has to pass "vertical line test". What is the verdict on that test result?
 

tkhunny

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I think that y is a function of x because graphing the function I get f(x)=−√x has the same solutions as y = −√x.. I'm not sure if this makes sense though.
Nope. I don't get it.

Function (roughly stated). A valid given value of x produces exactly one y-value. aka "The Vertical Line Test".

Does \(\displaystyle y = -\sqrt{x}\) pass?

As for the latter question, pick two value on the graph and see if \(\displaystyle |y|<x\). This may help you on your way.
 

clairestdean

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Graphing it, I don't think y=-√x passes.
 

topsquark

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Graphing it, I don't think y=-√x passes.
It doesn't have to exist for all x, though. Does it pass the vertical line test where sqrt{x} exists?

-Dan
 

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clairestdean

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No vertical line hits the graph once so it is a function
 

clairestdean

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The vertical line only hits once so it is a function.
 

pka

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Consider y = −√x. Is y a function of x? Why or why not?
That question is absolutely meaningless because it is incomplete.
\(\displaystyle y=-\sqrt{x}\) is a function on the set of non-negative real numbers.

\(\displaystyle y=-\sqrt{x}\) is not a function on the set of real numbers.
 

tkhunny

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\(\displaystyle y=-\sqrt{x}\) is not a function on the set of real numbers.
clairestdean, I suspect your algebra teacher will struggle with this. Press forward bravely.
 

clairestdean

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Well, that's the question I received. So would this answer make sense? Considering y = −√x. y is indeed a function of x. This is proven if we graph y=-√x and use the vertical line test. The vertical line test states that if a vertical line intersects the graph more than once, it's not a function, if it intersects once, it is a function. Graphing y=-√x no matter where we draw a vertical line, it won't intersected more than once so it is indeed a function.
 

pka

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Is it true that |y| is always less than x? Show that it is or give a counter example.
Please put different questions in different threads.
I assume that this question is about \(\displaystyle y=-\sqrt{x}\) is a function on the set of non-negative real numbers.
So that you ask in that situation is it the case that \(\displaystyle |y|\le x~?\)
So consider \(\displaystyle x=\tfrac{1}{4}\) then so that \(\displaystyle y=-\sqrt{\tfrac{1}{4}}\)
But that means \(\displaystyle |y|=\tfrac{1}{2}\). You answer your own question.
Is \(\displaystyle |y|\le x~?\)
 

Jomo

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Here is the test I use. Imagine you take an x value in the domain which is (I am assuming but YOU really should state) the non-negative real numbers. If you take the square root of one of those numbers how many results do you get? Then put a negative sign in front of it. How many results do you get? If you can't get two or more results then it is a function.
 

pka

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\(\displaystyle y=-\sqrt{x}\) is not a function on the set of real numbers.
clairestdean, I suspect your algebra teacher will struggle with this. Press forward bravely.
The statement that \(\displaystyle f\) is an function from set \(\displaystyle A\) to set \(\displaystyle B\) means:
1) \(\displaystyle f\subseteq A\times B\)
2) For \(\displaystyle \forall s\in A\) then \(\displaystyle \exists t\in B\) such that \(\displaystyle (s,t)\in f\) we write \(\displaystyle f(s)=t\).
3) No two pairs in \(\displaystyle f\) have the same first term. That is If \(\displaystyle (r,g)\in f~\&~(r,h)\in f\) then it must be true that \(\displaystyle g=h\).
 
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