Problem 12 - Advanced Calculus

Question

Given \(E(x, y)=x^{2}-y^{2}\), how many different functions \(f\) are there that are "defined by the equation \(E(x, y)=0\) so that \(y=f(x) "\) ?

Step-by-Step Solution

Answer

There are two distinct functions defined by the equation \(E(x, y)=0\), namely \(f_1(x) = x\) and \(f_2(x) = -x\).

Step 1: Rewrite the Equation

First, express the equation \(E(x, y)=0\) which turns out to be \(x^{2}-y^{2}=0\).

Step 2: Solve for \(y\)

Next, solve for \(y\) to get the \(y\) in terms of \(x\). This can be done by adding \(y^{2}\) to each side of the equation to isolate \(y\). This will result in two possibilities: \(y=x\) and \(y=-x\). So actually the equation yields two separate functions.

Step 3: Identify Functions

Now that we have derived \(y\) in terms of \(x\), we have to stress that each expression corresponds to a distinct function, giving us two functions in total: \(f_1(x) = x\) and \(f_2(x) = -x\).

Key Concepts

Implicit Function Definition

An implicit function is not stated directly, but rather exists within an equation that relates two or more variables. Instead of being expressed as a clear formula like y = f(x), an implicit function might be interwoven within an equation where the dependent variable, typically y, is not isolated on one side.

For instance, in our exercise, the implicit function is given by the equation E(x, y) = 0, which is expanded to (x^2 - y^2) = 0. Here, the relationship between x and y is not direct, and y is not initially subject alone, representing an implicit function within the equation.

Solving Equations for Functions

When solving equations for functions, the goal is to express one variable explicitly in terms of another. This conversion from an implicit to an explicit relationship is beneficial for understanding how changes in one variable affect the other.

In practice, as we did in our exercise, you might rearrange the equation to isolate the variable of interest on one side. This process can involve a series of algebraic manipulations - like factoring, distributing, adding or subtracting terms - to solve for the variable.

Once ( y^2 ) is added to both sides of the equation, we end up with ( y^2 = x^2 ) which further simplifies to two potential functions:

• ( y = x ) (where ( y ) directly equals ( x )), and
• ( y = -x ) (where ( y ) is the additive inverse of ( x )).

Algebraic Functions

Algebraic functions are mathematical expressions that involve combinations of constants, variables and the arithmetic operations: addition, subtraction, multiplication, division, and root extraction.

The solutions from our exercise, f_1(x) = x and f_2(x) = -x, are examples of simple algebraic functions. They show a clear relationship between y, the dependent variable, and x, the independent variable, in the form of linear equations.

These functions are foundational in algebra and provide a stepping stone to more complex functions. They allow us to graph linear relationships easily and form a basis for understanding how variables can interact within an equation.