/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Determine whether the equation d... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the equation defines \(y\) as a function of \(x\) or defines \(x\) as a function of \(y\) $$x^{2}+2 x y+y^{2}=0$$

Short Answer

Expert verified
Answer: The given equation defines both \(y\) as a function of \(x\) and \(x\) as a function of \(y\), as it can be written as \(y=-x\) and \(x=-y\).

Step by step solution

01

Check if we can solve for y as a function of x

We will attempt to rewrite the given equation in terms of \(y\) as a function of \(x\). Start by trying to isolate \(y\) on one side of the equation. Rearrange the given equation: $$y^2 + 2xy + x^2 = 0$$ We can rewrite it as a "quadratic equation" in terms of \(y\): $$y^2 + 2xy + x^2 = (y + x)^{2} = 0$$ Now, if we can solve for \(y\), it should be a function of \(x\). We can solve for \(y\): $$y + x = 0 \Rightarrow y = -x$$ Here, \(y\) is expressed as a function of \(x\).
02

Check if we can solve for x as a function of y

Now, let's attempt to rewrite the equation in terms of \(x\) as a function of \(y\). Rearrange the given equation: $$x^2 + 2xy + y^2 = (x+y)^2 = 0$$ Now, if we can solve for \(x\), it should be a function of \(y\). We already have \(y = -x\). So, $$x = -y$$ Here, \(x\) is expressed as a function of \(y\).
03

Conclusion

The given equation $$x^{2}+2 x y+y^{2}=0$$ defines both \(y\) as a function of \(x\) and \(x\) as a function of \(y\). The equation can be written as \(y=-x\) and \(x=-y\), which shows the relationship between the two variables.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Defining Functions
In precalculus, a function is a special relationship between two sets where each element of the first set, called the domain, is paired with exactly one element of the second set, known as the range. When determining if an equation defines a function, the key is to check if for every input value there is only one output.

Consider the equation from our exercise: \(x^2 + 2xy + y^2 = 0\). By factoring, we recognize it as \( (y + x)^2 = 0\), which implies that \(y = -x\). Here, for every value of \(x\), there is a unique corresponding value of \(y\), which satisfies the definition of a function. This concept is essential because understanding when an equation defines a function is the foundation for exploring more complex function behavior and their graphs.
Quadratic Equations
A quadratic equation is a second-degree polynomial equation in a single variable \(x\), with the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The equation we have, \(y^2 + 2xy + x^2 = 0\), resembles a quadratic equation in terms of \(y\) when we treat \(x\) as a constant.

Quadratics are known for their characteristic 'U'-shaped curve called a parabola when graphed. In this exercise, the equation simplifies to \( (y + x)^2 = 0\), which actually represents a straight line, showing that even though the form is quadratic, it doesn't always yield a parabola.
Variable Isolation
Isolating a variable is a crucial step in solving equations, making one particular variable the subject of the formula. The goal is to 'untangle' the variable from other terms so that it stands alone on one side of the equation.

In our example, isolating \(y\) involves recognizing that \(y^2 + 2xy + x^2\) can be factored into \( (y + x)^2 \). Setting this equal to zero and taking the square root gives us \(y + x = 0\), which easily isolates \(y\) as \(y = -x\). Similarly, we can isolate \(x\) to show that it also can be expressed as a function of \(y\), which is \(x = -y\).
Function Notation
Function notation is a streamlined way of representing functions, using \(f(x)\) instead of \(y\). It emphasizes the function as a machine that takes an input \(x\) and produces an output \(f(x)\).

In the context of the exercise, if we wanted to define \(y\) as a function of \(x\), we could write \(f(x) = -x\). This notation is beneficial when working with multiple functions, as it clearly distinguishes which function you're referring to and which variable is independent. When it comes to \(x\) as a function of \(y\), we can similarly write \(g(y) = -y\), with \(g\) being another function entirely.

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Most popular questions from this chapter

Give an example of two different functions \(f\) and \(g\) that have all of the following properties: $$ f(-1)=1=g(-1) \quad \text { and } \quad f(0)=0=g(0) $$ and \(\quad f(1)=1=g(1)\)

Use algebra to find the inverse of the given one-to-one function. $$f(x)=-x+1$$

Find the radius \(r\) and height \(h\) of a cylindrical can with a surface area of 60 square inches and the largest possible volume, as follows. (a) Write an equation for the volume \(V\) of the can in terms of \(r\) and \(h\). (b) Write an equation in \(r\) and \(h\) that expresses the fact that the surface area of the can is \(60 .\) [ Hint: Think of cutting the top and bottom off the can; then cut the side of the can lengthwise and roll it out flat; it's now a rectangle. The surface area is the area of the top and bottom plus the area of this rectangle. The length of the rectangle is the same as the circumference of the original can (why?).] (c) Write an equation that expresses \(V\) as a function of \(r\) [Hint: Solve the equation in part (b) for \(h\), and substitute the result in the equation of part (a).] (d) Graph the function in part (c), and find the value of \(r\) that produces the largest possible value of \(V\). What is \(h\) in this case?

Find the approximate intervals on which the function is increasing, those on which it is decreasing, and those on which it is constant. $$f(x)=\frac{1}{x}$$

Verify that \((f \circ g)(x)=x\) and \((g \circ f)(x)=x\) for every \(x\) $$f(x)=9 x+8, \quad g(x)=\frac{x-8}{9}$$

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