# 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

##### New member
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

##### Super Moderator
Staff member
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 follow the rules of posting in this forum, as enunciated at:

#### clairestdean

##### New member
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

##### Super Moderator
Staff member
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

##### Moderator
Staff member
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

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

#### tkhunny

##### Moderator
Staff member
Graphing it, I don't think y=-√x passes.
Provide an example of not passing.

#### topsquark

##### Full Member
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

#### Attachments

• 12.7 KB Views: 2

#### clairestdean

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

#### clairestdean

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

#### pka

##### Elite Member
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

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

#### clairestdean

##### New member
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

##### Elite Member
Is it true that |y| is always less than x? Show that it is or give a counter example.
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

##### Elite Member
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

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