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Implicitly Defined Functions

Tom Green

Tom Green

7 min read

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Study Guide Overview

This study guide covers implicitly defined functions, contrasting them with explicit functions. It explains how to graph and solve for variables in implicit equations, emphasizing domain and range. The guide also discusses interpreting slope (positive, negative, zero, undefined) and identifying horizontal/vertical intervals. Finally, it provides practice questions covering these concepts.

Implicitly Defined Functions: Your Night-Before-the-Exam Guide

Hey there, future AP Pre-Calculus master! Let's dive into implicitly defined functions. Think of this as your cheat sheet for tonight – clear, concise, and ready to boost your confidence. Let's get started!

What are Implicitly Defined Functions?

An equation with both 'x' and 'y' can secretly describe one or more functions. The graph of such an equation is simply all the (x, y) pairs that make the equation true. 🧑‍🎨

Key Concept

Graphing Equations with x and y

  • Explicit Functions: You're used to these! Like y = 2x + 1. Plug in 'x', get 'y'. Graph is a line. ↗️

    Graph of the function y = 2x + 1

    • Caption: The graph of the function y = 2x + 1, a simple linear equation.

  • Implicit Functions: Equations where 'x' and 'y' are mixed, like x² + y² = 1. You can't easily isolate 'y'. 💡

    Graph of the function x^2 + y^2 = 1

    • Caption: The graph of the function x² + y² = 1, a circle centered at the origin.
Memory Aid

Think of it like this: Explicit functions are like having a recipe where you directly get the output (y) from the input (x). Implicit functions are like a puzzle where you have to figure out the relationship between x and y to find the solutions. 🧩

Solving for One Variable

  • Solving for 'y' can give you one or more functions, each representing a piece of the original graph. For x² + y² = 1, solving for 'y' gives y = ±√(1 - x²), two functions (top and bottom halves of the circle). 〽️
  • Solving for 'x' can give you one or more functions, each representing a piece of the original graph. For x² + y² = 1, solving for 'x' gives x = ±√(1 - y²), two functions (left and right halves of the circle).
  • Domain & Range: Solving for one variable can reveal the domain and range of the resulting function. Be careful! The domain of the function might be more restricted than the domain of the original equation. 😮

Relating Functions Involving x and y

The graph of an implicitly defined function is a set of all ordered pairs (x, y) that satisfy the equation. 👌

  • Slope: The ratio of change in 'y' to the change in 'x' (Δy/Δx) between two very close points on the graph shows the relationship between the variables.

    • Positive Slope ↗️: As 'x' increases, 'y' also increases.
    • Negative Slope ↘️: As 'x' increases, 'y' decreases.
Memory Aid

Think of the slope like a hill: positive slope means you're going uphill, and negative slope means you're going downhill as you move from left to right on the graph.

Special Cases: Horizontal and Vertical Intervals

  • Horizontal Interval ↔️: When the rate of change of x with respect to y is zero (Δx/Δy = 0), the graph moves horizontally. The slope is zero.
  • Vertical Interval ↕️: When the rate of change of y with respect to x is zero (Δy/Δx is undefined), the graph moves vertically. The slope is undefined.
Common Mistake

Don't confuse zero slope (horizontal line) with undefined slope (vertical line). Remember, a vertical line has an infinite slope.

Example

Let's revisit x² + y² = 1. If we take two points (0.8, 0.6) and (0.9, 0.7) which are close together, the rate of change of x and y is (0.9 - 0.8) / (0.7 - 0.6) = 1, the x and y values increase simultaneously. ✅

If we take two points (0.8, 0.6) and (0.8, 0.7) which are close together, the rate of change of x with respect to y is (0.8 - 0.8) / (0.7 - 0.6) = 0, indicating a horizontal interval. ✅

Final Exam Focus

Okay, deep breaths! Here's what to focus on for the exam:

  • Implicit vs. Explicit: Know the difference and how to switch between them.
  • Solving for Variables: Practice isolating 'x' or 'y' to find functions.
  • Domain and Range: Don't forget to consider restrictions when solving for a variable.
  • Slope Interpretation: Understand how the slope (positive, negative, zero, undefined) relates to the graph's behavior.
  • Horizontal and Vertical Intervals: Recognize when the rate of change is zero or undefined.
Exam Tip

When you see an implicit equation, ask yourself: Can I solve for 'y'? If not, how can I analyze the relationship between 'x' and 'y' based on the slope?

Last-Minute Tips

  • Time Management: Don't get stuck on one question. Move on and come back if needed.
  • Common Pitfalls: Watch out for domain restrictions and confusing zero vs. undefined slopes.
  • Challenging Questions: Break down complex problems into smaller, manageable parts.

Practice Questions

Practice Question

Multiple Choice Questions

  1. Which of the following equations implicitly defines a function where y is not a function of x? (A) y=x2+3y = x^2 + 3 (B) x=y2x = y^2 (C) y=xy = \sqrt{x} (D) y=2x−5y = 2x - 5

  2. The equation x2+y2=25x^2 + y^2 = 25 represents a circle. What are the functions obtained when solving for y? (A) y=25−x2y = \sqrt{25 - x^2} and y=−25−x2y = -\sqrt{25 - x^2} (B) y=25+x2y = \sqrt{25 + x^2} and y=−25+x2y = -\sqrt{25 + x^2} (C) y=5−xy = 5 - x and y=−5+xy = -5 + x (D) y=x2−25y = \sqrt{x^2 - 25} and y=−x2−25y = -\sqrt{x^2 - 25}

  3. If the rate of change of x with respect to y is zero, what does this imply about the graph of the function? (A) The graph is moving vertically. (B) The graph is moving horizontally. (C) The graph has a positive slope. (D) The graph has a negative slope.

Free Response Question

Consider the equation x2+4y2=16x^2 + 4y^2 = 16.

(a) Sketch the graph of the equation.

(b) Solve the equation for y, expressing y as a function of x.

(c) Determine the domain and range of the function(s) found in part (b).

(d) Find the slope of the line tangent to the graph at the point (2, \sqrt{3}).

Scoring Breakdown:

(a) (2 points): 1 point for correct shape (ellipse), 1 point for correct intercepts.

(b) (2 points): 1 point for correct algebraic manipulation, 1 point for expressing y as a function of x (including both positive and negative roots).

(c) (2 points): 1 point for correct domain, 1 point for correct range.

(d) (3 points): 1 point for implicit differentiation, 1 point for correct substitution, 1 point for correct slope.

Answers

Multiple Choice:

  1. (B)
  2. (A)
  3. (B)

Free Response:

(a) The graph is an ellipse centered at the origin, with x-intercepts at ±4 and y-intercepts at ±2. (b) Solving for y:

4y2=16−x24y^2 = 16 - x^2

y2=4−x24y^2 = 4 - \frac{x^2}{4}

y=±4−x24y = ±\sqrt{4 - \frac{x^2}{4}}

y=±1216−x2y = ±\frac{1}{2}\sqrt{16 - x^2}

(c) Domain: −4≤x≤4-4 \le x \le 4, Range: −2≤y≤2-2 \le y \le 2

(d) Implicit differentiation:

2x+8ydydx=02x + 8y \frac{dy}{dx} = 0

dydx=−2x8y=−x4y\frac{dy}{dx} = -\frac{2x}{8y} = -\frac{x}{4y}

At (2, \sqrt{3}):

dydx=−243=−123=−36\frac{dy}{dx} = -\frac{2}{4\sqrt{3}} = -\frac{1}{2\sqrt{3}} = -\frac{\sqrt{3}}{6}

You've got this! You're well-prepared and ready to rock the AP Pre-Calculus exam. Good luck!

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Question 1 of 8

Which of the following equations represents an explicit function? 🤔

x2+y2=4x^2 + y^2 = 4

y=3x−7y = 3x - 7

x=y2+1x = y^2 + 1

xy=5xy = 5