Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

Which of the following determine a function \(f\) with formula \(y=f(x) ?\) For those that do, find \(f(x)\). Hint: Solve for \(y\) in terms of \(x\) and note that the definition of a function requires a single \(y\) for each \(x\). (a) \(x^{2}+y^{2}=1\) (b) \(x y+y+x=1, x \neq-1\) (c) \(x=\sqrt{2 y+1}\) (d) \(x=\frac{y}{y+1}\)

Step-by-Step Solution

Answer

Functions are (b) \(f(x) = \frac{1-x}{x+1}\), (c) \(f(x) = \frac{x^2-1}{2}\), and (d) \(f(x) = \frac{x}{1-x}\).

Step 1: Analyze equation (a)

Equation (a) is \(x^{2} + y^{2} = 1\). This represents a circle centered at the origin with radius 1. To solve for \(y\) in terms of \(x\), we isolate \(y\): \(y^{2} = 1 - x^{2}\), then take the square root to get \(y = \pm \sqrt{1 - x^{2}}\). Since we have both \(+\) and \(-\) solutions, for the same \(x\) there are two possible \(y\) values, which violates the definition of a function.

Step 2: Analyze equation (b)

Equation (b) is \(xy + y + x = 1\). Rearrange to find a common factor: \(y(x+1) + x = 1\), which simplifies to \(y = \frac{1-x}{x+1}\) given \(x eq -1\). For each \(x\) (except \(-1\)), there is a unique \(y\), satisfying the definition of a function. Thus, \(f(x) = \frac{1-x}{x+1}\) is a valid function.

Step 3: Analyze equation (c)

Equation (c) is \(x = \sqrt{2y + 1}\). Square both sides to solve for \(y\): \(x^2 = 2y + 1\), which simplifies to \(y = \frac{x^2 - 1}{2}\). Each \(x\) yields one unique \(y\), so it satisfies the definition of a function. The function is \(f(x) = \frac{x^2 - 1}{2}\).

Step 4: Analyze equation (d)

Equation (d) is \(x = \frac{y}{y+1}\). To express \(y\) in terms of \(x\), cross-multiply: \(x(y+1) = y\) gives \(xy + x = y\). Rearrange to get \(y = \frac{x}{1-x}\). It doesn't allow multiple \(y\) values for one \(x\) because division by \(1-x\) doesn't introduce any additional \(y\) solutions. Thus, \(f(x) = \frac{x}{1-x}\) is a function.

Key Concepts

Functions

In calculus, one of the fundamental ideas is the concept of a function. A function is a rule or a relation that uniquely assigns exactly one output number, often denoted as \(y\), for each possible input number \(x\). This unique assignment is crucial; if there are two different \(y\) values for the same \(x\), it's not a function. For example, the equation \(x^2 + y^2 = 1\) represents a circle and does not define \(y\) as a function of \(x\) because for some \(x\) values, there are two possible \(y\) values.

Solving Equations

Solving equations involves finding the value(s) of a variable that make the equation true. In the context of determining functions, solving for \(y\) in terms of \(x\) is crucial. The objective is to isolate \(y\) on one side of the equation, expressing it solely as a function of \(x\). For instance, in the equation \(xy + y + x = 1\), rearranging terms to isolate \(y\) results in \(y = \frac{1-x}{x+1}\). Once isolated, one needs to verify that each \(x\) has a single corresponding \(y\). This ensures the relation is a function.

Function Definition

The definition of a function is based on the principle that each input should have only one output. It is like a machine where you put in a number \(x\) and get out a uniquely determined number \(y\). A classic example is the function represented by a graph: each \(x\) should correspond to exactly one \(y\) value. This is often described by the vertical line test, where a vertical line drawn anywhere along the graph of a function will intersect it at no more than one point. This definition helps to distinguish functional relationships from those that aren't functions.

Algebraic Manipulation

Algebraic manipulation is a set of techniques used to simplify or rearrange equations. This includes operations such as factoring, expanding, or distributing to isolate a variable. When identifying functions, you often need to manipulate an equation to express one variable in terms of another. For instance, from the expression \(x = \frac{y}{y+1}\), you can perform algebraic operations such as cross-multiplying and rearranging terms to express \(y\) as \(y = \frac{x}{1-x}\). This skill is crucial for transforming equations so that you can analyze them as functions.