Is this equation a function?

Question

I was having trouble algebraically verifying that this equation was a function.

$$x^2y - x^2 + 4y = 0.$$

I tried simplifying it like this:

$$x^2y - x^2 + 4y = 0.$$

$$x^2(y-1) = -4y.$$

$$x^2= \frac{-4y}{y-1}.$$

I don't think thats the best way of representing it so I just plugged in values into the initial equation. I think that it is a function for all real numbers except 0. Is this correct? Was there a better way of demonstrating it?

An equation is never a function. But an equation (in two variables) may determine one variable as a function of the other. However, the answer may very well depend on which variable you consider to be the independent one. In other words: your question needs more detail/context. In the above you could just as well ask whether the equation determines $y$ as a function of $x$. You have investigated if it determines $x$ as a function of $y$. — mrf, Sep 22 '15 at 20:48

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

1

Is the equation a function?

$$x^2y - x^2 + 4y = 0.$$

$$x^2(y-1) = -4y.$$

$$x^2= \frac{-4y}{y-1}.$$

$$x= \pm \sqrt{\frac{4y}{1-y}}.$$

No, the square root is multivalued. For instance,

$$(-4)^2=4^2$$

See wolfram's plot.

However, if you solve for $y$, we get,

$$y={{x^2} \over {x^2+4}}$$

Which is a function.

edited Jun 12 '20 at 10:38

Community

1

answered Sep 22 '15 at 19:41

Zach466920

8,341

R.N · Answer 2 · 2015-09-22T20:31:16.403

0

$x^2y - x^2 + 4y = 0.$ thus $y(x^2+1)=x^2$ that gives $y=\frac{x^2}{x^2+1}$, for showing that is function Let $x_1=x_2$ then it is obvious that $y_1=y_2$ finished

edited Sep 22 '15 at 20:31

answered Sep 22 '15 at 19:36

R.N

4,318

My apologies I left out a y in the first part of the equation. – Joshua Morris Sep 22 '15 at 19:37
it is ok, dear. I edit my answer. – R.N Sep 22 '15 at 19:40
What is this? I need a more detailed explanation. How is this a proof? – Zach466920 Sep 22 '15 at 20:07
dear @Zach466920 it is just same as second part of your answer. I am agree with you that your answer is better. – R.N Sep 22 '15 at 20:19
I down vote because you purport to have a proof, but in the end it is exactly what my conclusion is except you claim to have proof, which is invalid. – Zach466920 Sep 22 '15 at 20:23

Is this equation a function?

2 Answers2