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$$F\left(\frac{1} {x}\right)=x^3 - 2$$

$$F(x)=\left(\frac{1} {x}\right)^3 - 2$$

$$F\left(\frac{x} {y}\right) = \left(\frac{y} {x}\right)^3 - 2 $$

I don't understand what I'm doing wrong.

Martin Sleziak
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Myles
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    Why do you think there is something wrong? – copper.hat Sep 11 '13 at 07:27
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    Hear, hear! Looks fine to me, just so $x \ne 0$! – Robert Lewis Sep 11 '13 at 07:30
  • @RobertLewis It would also be correct for x=0. Why should it be wrong?, also 0! = 1 :P – Daniel Sep 11 '13 at 11:42
  • @simpleBob You can't divide by zero. – user37238 Sep 11 '13 at 11:49
  • @user37238 Division by zero may be undefined (depending on the mathematical setting), but it is not incorrect. – Daniel Sep 11 '13 at 11:57
  • And an answer with undefined object is correct ? – user37238 Sep 11 '13 at 11:59
  • @user37238 If I define a function as f(x)=x/0, what would be f(5)? – Daniel Sep 11 '13 at 12:00
  • You said yourself that division by zero is undefined so how can you define $f$ ? – user37238 Sep 11 '13 at 12:03
  • @user37238 I also can define functions with undefined results. For example, division would be one of those functions. f(x,y)=x/y may be undefined for y=0 – Daniel Sep 11 '13 at 12:06
  • Then the domain of definition of $f$ is not $\mathbb{R}^2$ since you can't provide the image of (for example) $(1,0)$ by $f$. May be you should go see what is a function. – user37238 Sep 11 '13 at 12:15
  • In mathematics, you should not assume definitions that are not given. The question does not define the function to be f:ℝ->ℝ. And it obviously has a value that is undefined, that does not make the result be wrong. Note that the result of this function may not stay that way, the function may be transformed and the undefined result may suddenly become a well-formed result. – Daniel Sep 11 '13 at 13:33

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Everything is correct, except one thing: you should first make the assumption that $x\neq0$ since division by zero is undefined: $$F(x)=\left(\frac{1} {x}\right)^3 - 2\quad\text{where } x\neq0$$ And the other assumption is that in $\dfrac xy$, $y\neq 0$ so that the following holds: (of course, we are always assuming that $x\neq 0$ since we are taking its inverse) $$F\left(\dfrac xy\right)=\left(\dfrac yx\right)^3-2\quad\text{where $x\neq0$ and $y\neq0$ }.$$ Since you could easily derive nonsense without those important assumptions, example: $$F(1/0)=-2\text{ and }F(2/0)=-2\\\text{ $\,$so: }(0/1)^3-2=(0/2)^3-2 \\ (0/1)^3=(0/2)^3 \\ (0/1)^{-3}=(0/2)^{-3}\\ 1=2$$