Question

hey, if x=y^3, then x^(1/3)=y....right?

Answer 1

Yes!

Answer 2

Welcome to OpenStudy!

Answer 3

then, why is the graph of x=y^3 different from y=x^(1/3).... is math broken or what?

Answer 4

because the graph of the right does not include the negative side...

Answer 5

I know, but why is that?

Answer 6

isn't math correctly axiomatizised?

Answer 7

axio-what? um...im in calc, but I have no idea what that means

Answer 8

it means that a long time ago they were established together in one only science, and base themselves in certain principles, the fact that one of the "opposite operations" doesnt work the way is intended would mean that that was not met, it would be like saying that substracting is just the opposite of adding when you substract a positive number

Answer 9

Smartie-pants (;

Answer 10

Essentially, the function in your software package, "raised to the 1/3 power" is implicitly expecting only real inputs and is as such not defined for negative inputs. We have other functions in the complex realm which are defined for negative inputs and thus the two graphs are identical provided you are using software which allows for this.

Answer 11

Salvador: You ask good questions!

Answer 12

graph them by hand in \(\mathbb{R}^2\).

Answer 13

Since \((-1)^{1/3}=-1\), then \((-1,-1)\) should be a point on both graphs as well as \((0,0)\). The fact that your calculator doesn't graph that only speaks to your calculator.

Also, not all functions have function inverses.

Answer 14

\[x=y^{3}\]
taking log on both sides,
log x= 3 log y
therefore,
\[\frac{ 1 }{ 3}\log x= \log y\]
taking antilog on both sides,
\[x ^{\frac{ 1 }{ 3}}=y\]
so you are right.....

Answer 15

so it's just the calculator then?

Answer 16

thank you guys, its a lot of fun to see your explanations I'm closing this thing