Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.1.55

Chapter 7, Problem 7.1.55

Show that increasing functions and decreasing functions are one-to-one. That is, show that for any x₁ and x₂ in I, x₂ ≠ x₁ implies f(x₂) ≠ f(x₁).

Verified step by step guidance

Recall the definition of a one-to-one (injective) function: a function \( f \) is one-to-one if for any \( x_1, x_2 \) in the domain, \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). Equivalently, if \( x_1 \neq x_2 \), then \( f(x_1) \neq f(x_2) \).

Consider \( f \) to be an increasing function on an interval \( I \). By definition, for any \( x_1, x_2 \in I \) with \( x_1 < x_2 \), we have \( f(x_1) < f(x_2) \).

To show \( f \) is one-to-one, assume \( x_1 \neq x_2 \). Without loss of generality, suppose \( x_1 < x_2 \). Since \( f \) is increasing, it follows that \( f(x_1) < f(x_2) \), so \( f(x_1) \neq f(x_2) \).

Similarly, if \( f \) is decreasing on \( I \), then for any \( x_1 < x_2 \), \( f(x_1) > f(x_2) \). Using the same reasoning, \( x_1 \neq x_2 \) implies \( f(x_1) \neq f(x_2) \), so \( f \) is one-to-one.

Thus, both increasing and decreasing functions satisfy the condition that distinct inputs map to distinct outputs, proving they are one-to-one functions.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Monotonic Functions

A function is monotonic if it is either entirely non-increasing or non-decreasing over its domain. Increasing functions satisfy f(x₂) > f(x₁) whenever x₂ > x₁, while decreasing functions satisfy f(x₂) < f(x₁) for x₂ > x₁. This property ensures a consistent order in function values.

One-to-One (Injective) Functions

A function is one-to-one if different inputs produce different outputs, meaning f(x₁) = f(x₂) implies x₁ = x₂. This property guarantees that the function has an inverse on its range, as no two distinct domain points map to the same value.

Proof Using Contradiction for Injectivity

To prove monotonic functions are one-to-one, assume the opposite: that two distinct inputs map to the same output. Using the monotonicity property, this leads to a contradiction because increasing or decreasing behavior prevents equal outputs for different inputs, confirming injectivity.

Recommended video:

08:14

Using The Acceleration Function

Related Practice

Textbook Question

Use l’Hôpital’s rule to find the limits in Exercises 7–52.

17. lim (θ → π/2) (2θ - π) / cos(2π - θ)

Textbook Question

19. Show that e^x grows faster as x→∞ than x^n for any positive integer n, even x^1,000,000. (Hint: What is the nth derivative of x^n?)

views

Textbook Question

Use l’Hôpital’s rule to find the limits in Exercises 7–52.

19. lim (θ → π/6) (sin θ - 1/2) / (θ - π/6)

Textbook Question

Evaluate the integrals in Exercises 97–110.

101. ∫ (log₁₀x / x) dx

views

Textbook Question

Suppose that the differentiable function y = f(x) has an inverse and that the graph of f passes through the point (2, 4) and has a slope of 1/3 there. Find the value of df⁻¹/dx at x = 4.

Textbook Question

84. Find lim(x→∞) (√(x² + 1) - √x).

views