Implicitly Defined Functions - AP PreCalculus Study Guide

The Core Idea: Implicitly Defined Functions

In mathematics, we often express relationships between two variables, $x$ and $y$, using an equation. Such an equation, like $x^2+y^2=1$ or $x+y=1$, defines a relation. The core idea of this topic is to investigate when these relations can also be considered functions. A function requires that for every valid input `x$, there is exactly one corresponding output $y$. Some relations, like $x+y=1$, naturally satisfy this rule because we can write it as $y=1-x$, which is a function.

However, many relations are not functions. For the relation $x^2+y^2=1$, an input of $x=0$ leads to two possible outputs: $y=1$ and $y=-1$. This violates the definition of a function. The central task is to take such a relation and restrict it to create one or more new, valid functions. For the circle, we can define two separate functions: one for the upper semi-circle ($y=\sqrt{1-x^2}$) and another for the lower semi-circle ($y=-\sqrt{1-x^2}$). By analyzing the defining equation of these new, implicitly defined functions, we can determine their specific domains and ranges.

Key Definitions and Concepts

This topic focuses on the conceptual understanding of relations and functions rather than a list of formulas. The key concepts are definitions that distinguish these ideas.

• Relation: A relation is a rule defined by an equation in two variables that connects inputs (typically $x$) to outputs (typically $y$).

  • Example: $x^2+y^2=1$ is a relation. It describes all points $(x,y)$ that are 1 unit from the origin.

• Function: A function is a specific type of relation where each valid input `x$ corresponds to exactly one output $y$.

  • Example: The relation $x+y=1$ is also a function because for any $x$ you choose, there is only one $y$ that satisfies the equation. We can write this explicitly as $y=f(x)=1-x$.

• Restricting a Relation: This is the process of limiting the possible output values of a relation to create a valid function. This is typically done by solving the relation's equation for $y$ and choosing either the positive or negative result of a square root (or other operation that can produce multiple values).

  • Example: For the relation $x^2+y^2=1$, solving for $y$ gives $y^2=1-x^2$, which leads to $y=\pm \sqrt{1-x^2}$.

    • The function $f(x)=\sqrt{1-x^2}$ is a restriction of the relation to non-negative $y$-values (the upper semi-circle).

    • The function $g(x)=-\sqrt{1-x^2}$ is a restriction of the relation to non-positive $y$-values (the lower semi-circle).

Understanding Relations vs. Functions

The critical nuance of this topic is the precise distinction between a relation and a function. While all functions are relations, not all relations are functions. The defining test is whether a single input can produce multiple outputs.

Consider the relation defined by the equation $x^2+y^2=1$. If we are asked to find the value(s) of $y$ when $x=0.6$, we would solve:

$$(0.6{)}^2+y^2=1$$

$$0.36+y^2=1$$

$$y^2=1-0.36$$

$$y^2=0.64$$

$$y=\pm \sqrt{0.64}$$

$$y=0.8\text{and}y=-0.8$$

Since the single input `x=0.6$ corresponds to two different outputs, $y=0.8$ and $y=-0.8$, the relation $x^2+y^2=1$ is not a function of $y$ in terms of $x$.

Now, consider the relation $x+y=1$. If we use the same input x=0.6, we solve:

$$0.6+y=1$$

$$y=1-0.6$$

$$y=0.4$$

Here, the input `x=0.6$ produces only one output, $y=0.4$. This holds true for any $x$ we choose. Therefore, the relation $x+y=1$ is a function.

The process of "restricting" a non-function relation is our method for creating well-behaved functions from it. By choosing $y=\sqrt{1-x^2}$, we are explicitly stating that we will only consider the non-negative output for any given $x$. With this restriction, $x=0.6$ now yields only $y=0.8$, satisfying the definition of a function.

Core Concepts & Rules

• A relation is any equation that involves two variables, such as $x$ and $y$.

• A relation qualifies as a function only if every input value $x$ in its domain maps to a single, unique output value $y$.

• To test if a relation is a function, one can attempt to solve for $yFormula[9]9(0{)}^2+4y^2=36Formula[10]0+4y^2=36Formula[11]y^2=9Formula[12]y=\pm 3Formula[13]9x^2+4y^2=36Formula[14]4y^2=36-9x^{2}Formula[15]y^2=\frac{36-9x^2}{4}Formula[16]y=\pm \sqrt{\frac{36-9x^2}{4}}Formula[17]y=\pm \frac{\sqrt{9(4-x^2)}}{2}Formula[18]y=\pm \frac{3\sqrt{4-x^2}}{2}Formula[19]4-x^2\ge 0Formula[20]4\ge x^{2}Formula[21]-2\le x\le 2Formula[22]y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}Formula[23]y=\frac{-4\pm \sqrt{4^2-4(1)(3-x)}}{2(1)}Formula[24]y=\frac{-4\pm \sqrt{16-12+4x}}{2}Formula[25]y=\frac{-4\pm \sqrt{4+4x}}{2}Formula[26]y=\frac{-4\pm \sqrt{4(1+x)}}{2}Formula[27]y=\frac{-4\pm 2\sqrt{1+x}}{2}Formula[28]y=-2\pm \sqrt{1+x}Formula[29]h(x)=-2+\sqrt{1+x}Formula[30]1+x\ge 0Formula[31]x\ge -1$$

  The domain of $h(x)$ is $[-1,\infty )$.

• Range:

  The range is determined by the output values of $h(x)=-2+\sqrt{1+x}$. We know from the problem's restriction that the range must be $y\ge -2$. Let's confirm this from the equation. The minimum value of $\sqrt{1+x}$ is $0$ (when $x=-1$).

  • Minimum $h(x)$ value: $-2+0=-2$.

  As $x$ increases, $\sqrt{1+x}$ increases without bound. Therefore, the range of $h(x)$ is $[-2,\infty )$.

Using Your Calculator

The concepts in this topic are primarily analytical, meaning they are solved using algebra rather than a calculator. A graphing calculator cannot directly graph an implicit relation like $x^2+y^2=1$ in its standard function (Y=) mode.

However, a calculator is an excellent tool for verifying your results. After you have analytically solved for the restricted functions, you can graph them to confirm their shape, domain, and range.

Example: Verifying the functions from $x^2+y^2=1$

1. Analytically solve for $y$: As shown previously, this gives $y=\sqrt{1-x^2}$ and $y=-\sqrt{1-x^2}$.

2. Enter the functions into the calculator:

   • Press the Y= button.

   • In $Y_1$, enter √(1 - X^2)

   • In $Y_2$, enter -√(1 - X^2)

3. Graph the functions:

   • Press the GRAPH button. You will see the upper and lower semi-circles, which together form a circle.

   • Tip: The graph may look like an ellipse. Use the ZOOM -> 5:ZSquare option to adjust the viewing window for a more accurate circular shape.

4. Verify Domain and Range:

   • Use the TRACE function. As you move the cursor along $Y_1$, you will see that the $x$-values are between -1 and 1, and the $y$-values are between 0 and 1. This visually confirms the domain $[-1,1]$ and range $[0,1]$ for the upper semi-circle.

   • Switch to $Y_2$ (by pressing the down arrow) and trace again. This will confirm the domain $[-1,1]$ and range [-1, 0]` for the lower semi-circle. ## AP Exam Quick Hit ### Common Question Types - **Creating a Function from a Relation:** Given a relation like $x = y^2 - 4, you might be asked to find the equation that defines $y$ as a function of $x$ for $y\ge 0$.

   • Example: Solving gives $y=\pm \sqrt{x+4}$. Since we require $y\ge 0$, the correct function is $f(x)=\sqrt{x+4}$.

• Determining the Domain or Range of a Restricted Function: You will be given a relation and a restriction, and asked to find the domain or range of the resulting function.

  • Example: "A function $g(x)$ is created by restricting the relation $x^2+(y-1{)}^2=9$ to $y\le 1$. What is the range of $g(x)$?"

  • Solution: The relation is a circle centered at $(0,1)$ with radius 3. Its $y$-values span from $1-3=-2$ to $1+3=4$. The restriction $y\le 1$ means we are considering the lower semi-circle. Therefore, the range is $[-2,1]$.

Common Mistakes

• Forgetting the $\pm$ When Taking a Square Root: When solving an equation like $y^2=16-x^2$ for $y$, a common mistake is to only write $y=\sqrt{16-x^2}$ and forget the negative solution, $y=-\sqrt{16-x^2}$. This causes you to miss one of the possible restricted functions.

• Algebraic Errors in Isolating $y$: For more complex relations, such as $x=y^2-6y+5$, students may make errors when using the quadratic formula or completing the square to solve for $y$.

• Incorrectly Determining the Domain: After finding a function like $f(x)=\sqrt{x+4}$, a student might incorrectly set $x\ge 0$ out of habit, instead of correctly analyzing the radicand: $x+4\ge 0⟹x\ge -4$.

• Confusing Domain and Range: Students may correctly find the interval for the domain (e.g., $[-2,2]$) and then mistakenly apply the same interval to the range, without analyzing the output values of the function.

• Misinterpreting the Restriction: When given a restriction like $y\le 0$, students might correctly solve for $y=\pm \sqrt{...}$ but then incorrectly choose the positive function, $y=+\sqrt{...}$, instead of the negative one that satisfies the condition.