Textbook Question
Functions
In Exercises 1–6, find the domain and range of each function.
f(t) = 4/(3 − t)
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Ch. 1 - Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 1, Problem 1.1.14
Finding Formulas for Functions
Consider the point (x,y) lying on the graph of y = √(x − 3). Let L be the distance between the points (x,y) and (4,0). Write L as a function of y.
Verified step by step guidance1
Start by identifying the given point (x, y) on the graph of the function y = √(x − 3). This implies that x can be expressed in terms of y as x = y^2 + 3.
Recognize that the distance L between the points (x, y) and (4, 0) can be calculated using the distance formula: L = √((x - 4)^2 + (y - 0)^2).
Substitute the expression for x from step 1 into the distance formula: L = √(((y^2 + 3) - 4)^2 + y^2).
Simplify the expression inside the square root: ((y^2 + 3) - 4) simplifies to (y^2 - 1). Therefore, the expression becomes L = √((y^2 - 1)^2 + y^2).
Further simplify the expression: Expand (y^2 - 1)^2 to get y^4 - 2y^2 + 1. Thus, L = √(y^4 - 2y^2 + 1 + y^2), which simplifies to L = √(y^4 - y^2 + 1).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Distance Formula
The distance formula calculates the distance between two points in a Cartesian plane. For points (x1, y1) and (x2, y2), the distance L is given by L = √((x2 - x1)² + (y2 - y1)²). In this problem, we will use this formula to find the distance between the point (x, y) on the curve and the point (4, 0).
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5:04
Converting between Degrees & Radians
Function Representation
A function represents a relationship between two variables, typically denoted as y = f(x). In this context, we need to express the distance L as a function of y, which may involve substituting y into the distance formula and rearranging the equation to isolate L. This process is essential for understanding how L varies with y.
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Substitution in Functions
Substitution is a technique used in algebra and calculus to replace a variable with another expression. In this case, since y = √(x - 3), we can express x in terms of y by rearranging the equation to x = y² + 3. This substitution allows us to rewrite the distance L solely in terms of y, facilitating the analysis of the relationship between L and y.
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Related Practice
Textbook Question
Algebraic Combinations
In Exercises 3 and 4, find the domains of f, g, f/g and g/f.
f(x) = 1, g(x) = 1 + √x
Textbook Question
In Exercises 39–42, express the given quantity in terms of sin x and cos x.
cos (3π/2 + x)
Textbook Question
In Exercises 9–16, determine whether the function is even, odd, or neither.
𝔂 = x cos x
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Textbook Question
Finding Formulas for Functions
Express the side length of a square as a function of the length d of the square’s diagonal. Then express the area as a function of the diagonal length.
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Textbook Question
General Sine Curves
For
f(x) = A sin ((2π/B)(x – C) +D
identify A, B, C, and D for the sine functions in Exercises 67–70 and sketch their graphs.
y = ½ sin (πx – x) + ½
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