# Thread: express y as a function of x

1. ## express y as a function of x

if x-8y^3=0 express y as a function of x

2. ## Re: express y as a function of x

Originally Posted by RaiderOfCooies
if x-8y^3=0 express y as a function of x
$y=\dfrac{\sqrt[3]{x}}{2}$

3. ## Re: express y as a function of x

Thank you but do you mind explaining the steps? i can figure out squared but cubed throws me off.

4. ## Re: express y as a function of x

$x - 8y^3 =0$

add $8y^3$ to both sides of the equation ...

$x = 8y^3$

divide both sides by $8$ ...

$\dfrac{x}{8} = y^3$

take the cube root of both sides ...

$\sqrt[3]{\dfrac{x}{8}} = \sqrt[3]{y^3}$

$\dfrac{\sqrt[3]{x}}{2} = y$

5. ## Re: express y as a function of x

Originally Posted by RaiderOfCooies
Thank you but do you mind explaining the steps? i can figure out squared but cubed throws me off.
So you know that if $x= y^2$ you "invert" by using the square root: $y= \sqrt{x}= x^{1/2}$.

Well, if $x= y^3$ you "invert" by using the cube root: $y= \sqrt[3]{x}= x^{1/3}$.

6. ## Re: express y as a function of x

Originally Posted by HallsofIvy
So you know that if $x= y^2$ you "invert" by using the square root: $y= \sqrt{x}= x^{1/2}$.
But $x = y^2$ isn't a function, and it can't be rewritten as a function of x, which is the topic of the OP's thread.

$x = y^2, y \ge 0$ is a function, and it can be rewritten as $y= \sqrt{x}= x^{1/2}$.

7. ## Re: express y as a function of x

Originally Posted by greg1313
But $x = y^2$ isn't a function, and it can't be rewritten as a function of x, which is the topic of the OP's thread.

$x = y^2, y \ge 0$ is a function, and it can be rewritten as $y= \sqrt{x}= x^{1/2}$.
$x=y^2$ is a function of $y$. Perhaps you mean that it is not a function of $x$. Or perhaps the term you are looking for is that it is not a bijective function. Bijectivity is the actual requirement for a function to have an inverse.

8. ## Re: express y as a function of x

Originally Posted by HallsofIvy
So you know that if $x= y^2$ you "invert" by using the square root: $y= \sqrt{x}= x^{1/2}$.Well, if $x= y^3$ you "invert" by using the cube root: $y= \sqrt[3]{x}= x^{1/3}$.
Originally Posted by greg1313
But $x = y^2$ isn't a function, and it can't be rewritten as a function of x, which is the topic of the OP's thread.
$x = y^2, y \ge 0$ is a function, and it can be rewritten as $y= \sqrt{x}= x^{1/2}$.
Originally Posted by SlipEternal
$x=y^2$ is a function of $y$. Perhaps you mean that it is not a function of $x$. Or perhaps the term you are looking for is that it is not a bijective function. Bijectivity is the actual requirement for a function to have an inverse.
If we are having a disagreement about functions should we not have a clear notion what a function is?
That is a question foundations class.

Definition: Suppose that each of $A~\&~B$ is a nonempty set.
The statement that $f$ is a function from $A$ to $B$ means that:
\begin{align*}&\bullet~f\subseteq A\times B.\\&\bullet~\text{every }x\in A\text{ is the first term in exactly one pair in }f\\ \end{align*}

Now if $A=B=\mathbb{R}$ it should be clear that $f=\{(x,x^2)\}$ is a function and $g=\{(x^2,x)\}$ is not!

Also $f=\{(x,x^3)\}~\&~g=\{(x^3,x)\}$ are both functions.