2

$$x = 4y^2$$

This equation confuses me, how is $y$ not defined as a function of $x$?

I manipulated the function and it works just fine, but my book says "Explain why the given equation does not define $y$ as a function of $x$."

$$\frac{x}{4} = y^2 \tag1$$

$$y = \sqrt \frac{x}{4} \tag2$$

$$y = \frac {\sqrt x}{2} \tag3$$

Blue
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Intensed
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  • See the fifth in the list of [standard inverse functions] (https://en.wikipedia.org/wiki/Inverse_function#Standard_inverse_functions). – Jan-Magnus Økland Jan 30 '24 at 09:47
  • The key to your question is what a function is. We normally say a function is a mapping from an element of the domain (a number) to one and only one element of the codomain (another number). If you solve for x (what was given), any y leads to a single x. If you solve for y (what you did), any x leads to two y values. That's not an ordinary single-valued function, and for your purposes, it's not a function y(x). Any root is like that. If you like you can say it is a multi-valued function, but that's probably not what you're studying given the question. – RobinSparrow Jan 30 '24 at 13:43

3 Answers3

5

Each positive $x$ can have two possible values of $y$.

For example for $x=4$, $y$ can be $1$ or $-1$.

Do remember the negative solution as well when you solve $y$ in terms of $x$.

Siong Thye Goh
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2

Let's ignore the $4$ in the denominator for a moment, just to keep things simple and concise.

$x = y^2$ does not necessarily mean $y = \sqrt{x}$.

$\sqrt{x}$ refers ony to the positive root of $x$. We need to consider the negative root too.

So,

$$x = y^2 \,\,\, \Longrightarrow \,\,\, y = \pm \sqrt{x}$$

In your case above,

$$y = \frac{\pm \sqrt{x}}{2}$$

See the problem now? For every positive value of $x$, we have two distinct values of $y$. This violates the condition for $y$ to be a function of $x$.

Haris
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2

Both $(x,y) = (4, 1)$ and $(x,y) = (4, -1)$ satisfy the equation $x^2=4y^2$, therefore, $y$ cannot be a function of $x$ (because if $y$ is a function of $x$, then for each $x$, we can only have one corresponding value of $y$).

The mistake you made was that in your equations, equation $(1)$ implies equation $(2)$, but it is not equivalent to equation $(2)$. There are solutions to the first equation that are not solutions to the second one.

5xum
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