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Thread: express y as a function of x

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    Question express y as a function of x

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

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

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

    $\dfrac{\sqrt[3]{x}}{2} = y$
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    Re: express y as a function of x

    Quote Originally Posted by RaiderOfCooies View Post
    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}.
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    Re: express y as a function of x

    Quote Originally Posted by HallsofIvy View Post
    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}.
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    Re: express y as a function of x

    Quote Originally Posted by greg1313 View Post
    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.
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    Re: express y as a function of x

    Quote Originally Posted by HallsofIvy View Post
    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}.
    Quote Originally Posted by greg1313 View Post
    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}.
    Quote Originally Posted by SlipEternal View Post
    $x=y^2$ 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.
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