Functions of Several Variables

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SUMMARY

The discussion focuses on evaluating the function f(x,y) = x^3 + y^3 - 3x - 12y + 20 when one variable is known. Specifically, the user seeks to determine the value of y given x = -1 and a specific output value for the function. The conversation highlights the need to substitute known values into the function to isolate y, demonstrating that with a known z value (output of the function), it is possible to solve for y using algebraic manipulation.

PREREQUISITES
  • Understanding of multivariable functions
  • Basic algebraic manipulation skills
  • Familiarity with function notation
  • Knowledge of graphing in three dimensions
NEXT STEPS
  • Study methods for isolating variables in multivariable equations
  • Learn about implicit functions and their applications
  • Explore graphing techniques for three-dimensional functions
  • Investigate numerical methods for solving nonlinear equations
USEFUL FOR

Students in calculus, educators teaching multivariable functions, and anyone interested in understanding the evaluation of functions of several variables.

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My question is mostly down to staggering ignorance of basic notation. I have been unable to find a straightforward answer anywhere - so I assume it must be pretty basic.

I have a function: [tex]f(x,y)=x^3+y^3-3x-12y+20[/tex]

Given that I know a value of x, ie [tex]x=-1[/tex], how do I extract a value of y using this notation?

I know when [tex]f(x)=x^2[/tex] I can use substitute [tex]f(x)[/tex] with [tex]y[/tex], but I find myself stuck with what to substitute [tex]f(x,y)[/tex] with...
 
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In order to evaluate your f(x, y), you need two numbers. What you get out isn't a y value; it's a z value. To graph the function, you need three dimensions. If you know a z value and an x value, there's the possibility that you can solve for the value you don't know: y. For example, if you know that f(x, y) = 10 and that x = 1, you might be able to solve the equation 10 = 1^3 + y^3 -3(1) - 12y + 20 to get y.

Does that make sense?
 
Yes, it actually does. Thank you Mark.
 

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