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

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

Step 1: Understanding Functions

A function is a relation where each input (domain element) corresponds to exactly one output (range element). For an equation to represent a function, solving for \(y\) must yield a unique value for each \(x\).

Step 2: Analyzing Equation (a)

Consider the equation \(x^2 + y^2 = 1\). Solving for \(y\) gives \(y = \pm \sqrt{1 - x^2}\). Since there are two possible values of \(y\) for each \(x\), this equation does not define \(y\) as a function of \(x\).

Step 3: Analyzing Equation (b)

For \(xy + y + x = 1\), rewrite as \(y = \frac{1 - x}{x + 1}\), assuming \(x eq -1\) to prevent division by zero. Each \(x\) in this modified domain corresponds to a unique \(y\), thus defining a function \(f(x) = \frac{1-x}{x+1}\).

Step 4: Analyzing Equation (c)

Consider \(x = \sqrt{2y + 1}\). Solving for \(y\) gives \(y = \frac{x^2 - 1}{2}\). Since \(x\) maps to a single \(y\) value, this defines a function \(f(x) = \frac{x^2 - 1}{2}\).

Step 5: Analyzing Equation (d)

Given \(x = \frac{y}{y+1}\), expressing \(y\) in terms of \(x\) leads to the quadratic \(yx + x = y\). Solving it gives \(y = \frac{x}{1-x}\) assuming \(x eq 1\) to prevent division by zero. Each \(x\) defines a unique \(y\), hence a function \(f(x) = \frac{x}{1-x}\).

Key Concepts

Function Definition

A function is a special kind of relationship between two variables, usually `x` and `y`. In this relationship, each `x` value is linked to exactly one `y` value. It's like matching every student with one locker. This means that no matter what `x` you pick, there should be only one `y` value that pairs with it.
In mathematics, this is established by the vertical line test on a graph: if a vertical line crosses the graph of the equation at more than one point, then the relation is not a function. In algebraic terms, a function must be expressed with a clear and singular outcome for `y` when given `x`.
For example, if you have an expression like `y = 2x + 3`, for every `x` you input, you will get one and only one value of `y`.

Solving Equations

To figure out if an equation is a function, you often need to solve the equation for `y` in terms of `x`. This process involves rearranging and simplifying the equation until `y` stands alone on one side of the equation.

• Start with the equation as given. For instance, if you have `x^2 + y^2 = 1`, your goal is to express `y` alone.
• Apply algebraic techniques like factoring, expanding, or isolating the variable `y` depending on the structure of the equation.
• The result should show only one value for every `x` value.
  For example, solving `xy + y + x = 1` gives `y = (1-x)/(x+1)`, which is a function because it gives a singular `y` for every `x` not equal to `-1`.

In cases where there are multiple possibilities for `y`, such as `y = +/− sqrt(1-x^2)`, the relation does not define a function.

Unique Solutions

In the world of functions, unique solutions are key. This means for every `x`, there should be only one `y` paired with it. If multiple `y` values exist for a single `x`, it defies the function definition.
Consider `x = sqrt(2y + 1)`. Solving for `y` gives `y = (x^2 - 1)/2`. Here, for every `x` that you choose, you get only one corresponding `y`. This verifies that the equation represents a function.
Remember:

• If solving the equation gives a square root or an absolute value, check carefully. These often produce more than one outcome and need reassessment.
• For equations like `x = y/(y+1)`, ensure you handle denominators correctly to avoid errors in unique pairing.

Ensuring unique solutions involves careful algebraic manipulation to prevent multiple outcomes.

Domain and Range

The domain and range of a function describe where the function starts and ends on the graph. The domain is all the possible input values (or `x` values), while the range is all possible output values (`y` values). These aspects are crucial to correctly defining a function.
For an equation such as `y = (1-x)/(x+1)`, the domain is all numbers except `x = -1` (because this would cause division by zero). The range is all numbers that `y` can take when valid `x` values are plugged into the equation.
Understanding domains and ranges:

• Make sure to exclude values that would result in division by zero from the domain.
• Consider restrictions on square roots, as negative numbers aren't defined in the set of real numbers.
• Think about the logical limits of `y` given the range of `x`.

Clearly defining these aspects ensures you have a functioning equation that behaves predictably across numbers it's applied to.