x in terms of y versus y in terms of x

[stillofthenight]

Say you have y=x^2 and you wanted to write it in terms of y. I would solve for x and write it as f(y)=y^1/2. Does f(y)=y^1/2 mean ...to have the input as a number on the y axis and the output as a number on the x axis. So you would go up the y axis 4 and across x axis 2 which is just another way to refer to the same coordinate point in the xy plane? Is that correct way of thinking about it?

 

[lookagain]

Say you have y=x^2 and you wanted to write it in terms of > > y < < .

You mean write it in terms of x, correct?

\(\displaystyle y \ = \ x^2\)

\(\displaystyle \pm\sqrt{y \ } \ = \ x\)

\(\displaystyle x \ = \ \pm\sqrt{y \ }\)



Were you asking for something else instead?

 

[stillofthenight]

yes x=sqrt of y

Does that mean f(y)= sqrt y , which means go up 4 on the y axis and then across two on the x. That eventually leads you to the same point on the xy plane just as f(x)=x^2 does.

So y=x^2 and f(y)= sqrt y are equivalent just written in different terms?

 

[lookagain]

yes x=sqrt of y \(\displaystyle \ \ \ \ \) What I typed with the plus or minus sign also includes x = -sqrt(y).

Does that mean f(y)= sqrt y , which means go up 4 on the y axis and then across two on the x.

No. You asked about a function leading to a supposedly specific point. That isn't logical. Also, you did not give the point (supposedly the origin)

you're "going up 4 [from] on the y-axis" to get to the final point.

That eventually leads you to the same point on the xy plane just as f(x)=x^2 does.

So y=x^2 and f(y)= sqrt y are equivalent just written in different terms?

No. In the former, there is no restriction on the sign of x. In the second, if you mean x = sqrt(y), x can never be negative.

That's all I will post on this.

 

[JeffM]

yes x=sqrt of y

Does that mean f(y)= sqrt y , which means go up 4 on the y axis and then across two on the x. That eventually leads you to the same point on the xy plane just as f(x)=x^2 does.

So y=x^2 and f(y)= sqrt y are equivalent just written in different terms?

If x is real and \(\displaystyle y = x^2,\) then the range of y is all non-negative real numbers, and y is a function of x.

If y is a non-negative real number, then \(\displaystyle \sqrt{y} \ge 0.\)

In other words, these two formulations are not equivalent because the second does not apply to all values of x whereas the first does.

Although positive real numbers (except 0) have a positive and a negative square root, the radical symbol refers ONLY to a non-negative square root.

Now if you say \(\displaystyle x = \pm \sqrt{y},\)

then you do not have a proper function because the resultant is ambiguous for every value of y except 0.