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Ch. 1 - Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 1 - FunctionsProblema 1.3.43a
Capitolo 1, Problema 1.3.43a

Splitting up curves The unit circle x² + y² = 1  consists of four one-to-one functions, ƒ₁ (x), ƒ₂(x) , ƒ₃(x), and ƒ₄ (x)  (see figure) <IMAGE>.


a. Find the domain and a formula for each function.

Guida verificata passo dopo passo
1
Step 1: Recognize that the unit circle x^2 + y^2 = 1 can be split into four segments, each representing a one-to-one function. These segments correspond to the four quadrants of the circle.
Step 2: For each quadrant, determine the range of x-values (domain) and the corresponding y-values. The unit circle is symmetric, so consider the signs of x and y in each quadrant.
Step 3: In the first quadrant, both x and y are positive. The function f_1(x) can be expressed as y = sqrt(1 - x^2) with the domain 0 <= x <= 1.
Step 4: In the second quadrant, x is negative and y is positive. The function f_2(x) can be expressed as y = sqrt(1 - x^2) with the domain -1 <= x <= 0.
Step 5: In the third quadrant, both x and y are negative. The function f_3(x) can be expressed as y = -sqrt(1 - x^2) with the domain -1 <= x <= 0.

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Domain of a Function

The domain of a function refers to the set of all possible input values (x-values) for which the function is defined. For the unit circle equation x² + y² = 1, the domain must be determined by isolating y in terms of x, which leads to two distinct functions for the upper and lower halves of the circle. Understanding the domain is crucial for identifying valid inputs and ensuring the function behaves correctly within its defined limits.
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One-to-One Functions

A one-to-one function is a type of function where each output value is associated with exactly one input value, meaning no two different inputs produce the same output. In the context of the unit circle, splitting it into four one-to-one functions allows us to express the circle in a way that each function can be inverted. This property is essential for solving problems that require finding inverse functions or analyzing the behavior of the functions over their respective domains.
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Implicit vs. Explicit Functions

Implicit functions are defined by an equation involving both variables, such as x² + y² = 1, while explicit functions express one variable directly in terms of another, like y = f(x). To find the formulas for the functions ƒ₁(x), ƒ₂(x), ƒ₃(x), and ƒ₄(x) from the unit circle, one must manipulate the implicit equation to derive explicit forms for y. This distinction is vital for understanding how to work with different types of functions in calculus.
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Domanda del libro di testo

Inverse of composite functions


c. Explain why if g and h are one-to-one, the inverse of ƒ(x) = g(h(x)) exists.

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Domanda del libro di testo

Inverse of composite functions


b. Let g(x) = x² + 1 and h(x) = √x. Consider the composite function ƒ(x) = g(h(x)). Find ƒ⁻¹ directly and then express it in terms of g⁻¹ and h⁻¹

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Domanda del libro di testo

Splitting up curves The unit circle x² + y² = 1  consists of four one-to-one functions, ƒ₁ (x), ƒ₂(x) , ƒ₃(x), and ƒ₄ (x)  (see figure)<IMAGE>.


b. Find the inverse of each function and write it as y= ƒ⁻¹ (x)

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Domanda del libro di testo

Determine whether the following statements are true and give an explanation or counterexample.


If y= 3ˣ , then x = ³√y

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Domanda del libro di testo

Evaluate and simplify the difference quotients (f(x + h) - f(x)) / h and (f(x) - f(a)) / (x - a) for each function.

f(x) = x2 - 2x

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Domanda del libro di testo

Inverse of composite functions


a. Let g(x) = 2x + 3 and h(x) = x³. Consider the composite function ƒ(x) = g(h(x)). Find ƒ⁻¹ directly and then express it in terms of g⁻¹ and h⁻¹

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