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]

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.

 

[Deleted member 4993]

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]

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.

 

[Deleted member 4993]

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]

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 $$y = -\sqrt{x}$$ pass?

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

 

[clairestdean]

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

 

[tkhunny]

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

Provide an example of not passing.

 

[topsquark]

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

 

[clairestdean]

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

 

[clairestdean]

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

 

[pka]

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]

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

clairestdean, I suspect your algebra teacher will struggle with this. Press forward bravely.

 

[clairestdean]

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]

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 $\displaystyle |y|\le x~?$

 

[Steven G]

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]

$\displaystyle y=-\sqrt{x}$ 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 $\displaystyle g=h$.