Algebraic Equations with Multiple Variables

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The discussion centers on solving algebraic equations with multiple variables, specifically f(x)=(x^5)/(x^2+6) and f(x)=2x^3-4x. Participants express confusion about finding inverse functions for these polynomials and emphasize the importance of understanding the definition of inverse functions. There is a suggestion to clarify the equations using proper notation for better comprehension. Additionally, resources for learning about inverse functions are recommended. Clear communication and precise mathematical formatting are essential for resolving the queries presented.
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f(x)=(x^5)/x^2+6 and f(x)=2x^3-4x



I tried x=(y^5)/(y^2+6) then x(y^2+6)=y^5 but idk after and
y=2x^3-4x then idk what to do
 
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Are you asking what are the inverse functions for these polynomials? If so, do you know the exact definition of "inverse function"? And if you do, would you start by stating the inverse function if ##f(x) = x^5##. Does that give you enough clue how to proceed?

If you aren't sure of the definition, it is certainly going to be impossible to do these problems. So look it up on the internet. I'll bet it's crawling with websites that discuss inverse functions and give examples.
 
sammiyahc0 said:
f(x)=(x^5)/x^2+6 and f(x)=2x^3-4x



I tried x=(y^5)/(y^2+6) then x(y^2+6)=y^5 but idk after and
y=2x^3-4x then idk what to do

Your post doesn't make sense to me. Can you please type the full question? And please use parens in your equations, so that the order of precedence is unambiguous. Even better would be to type them in LaTex:

https://www.physicsforums.com/showthread.php?t=710433

:smile:
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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