/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Assume that \(f\) is continuous ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 61

Assume that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b) .\) Also assume that \(f(a)\) and \(f(b)\) have opposite signs and that \(f^{\prime} \neq 0\) between \(a\) and \(b .\) Show that \(f(x)=0\) exactly once between \(a\) and \(b\)

Short Answer

Expert verified
The function \( f(x) = 0 \) exactly once in \((a, b)\) by IVT and non-zero derivative condition.

Step by step solution

01

Understand the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that if a function \( f \) is continuous on a closed interval \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then \(f(x) = 0\) for at least one value \( x \) in \((a, b)\). This is because the continuous function must cross the x-axis to change from a negative value to a positive value or vice versa.
02

Apply IVT to Ensure a Zero Exists

Since \( f \) is continuous on \([a, b]\) and \(f(a)\) and \(f(b)\) are of opposite signs, the IVT guarantees that there exists at least one \( x \) in \((a, b)\) such that \( f(x) = 0 \).
03

Leveraging Rolle's Theorem to Show Uniqueness

Rolle's Theorem states that if a function is continuous on \([a, b]\), differentiable on \((a, b)\), and \(f(a) = f(b)\), there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).
04

Ensure No Additional Zeros with Given Derivative Condition

Since \(f'(x) eq 0\) for \( x \) in \((a, b)\), there cannot be any \( c \) such that \( f'(c) = 0 \). This means there are no turning points, reinforcing that the graph of \( f(x) \) crosses the x-axis at exactly one point.
05

Conclude With Uniqueness of Root

Due to the constraints on \( f'(x) \), \( f(x) = 0 \) not only exists at least once by IVT but exactly once in \((a, b)\). No possibility of another zero exists because otherwise \( f'(x) = 0 \) would occur, contradicting the initial condition.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity
Continuity is a foundational concept in calculus and plays a vital role in many theorems. A function is said to be continuous on a closed interval \([a, b]\) if there are no breaks, jumps, or holes in its graph within that interval. This means that for any value within \([a, b]\), the function approaches and reaches this value smoothly.
Here are some crucial points to understand about continuity:
  • At a point \(x = c\), a function \(f(x)\) is continuous if the limit as \(x\) approaches \(c\) from both directions equals \(f(c)\).
  • A function is continuous on an interval if it is continuous at every point in that interval.
  • Intermediate Value Theorem depends on this smooth, unbroken nature of continuous functions.
  • For the Intermediate Value Theorem to apply, continuity on the closed interval \([a, b]\) is essential.
If you imagine drawing the graph of a continuous function between two points without lifting the pen, you have understood continuity.
Differentiability
Differentiability refers to the existence of a derivative for a function at every point in an open interval. A function is differentiable at a point \(x = c\) if the function has a derivative at that point, meaning it has a defined tangent. Differentiability implies continuity, but continuity does not necessarily imply differentiability.
Key elements of differentiability include:
  • A function \(f\) is differentiable on an open interval \( (a, b)\) if a derivative exists at every point in this interval.
  • If \(f\) is differentiable on \( (a, b)\), it is also continuous there.
  • Points of non-differentiability include sharp corners or cusps.
  • Differentiability ensures that there are no abrupt changes in the slope of the function within the interval.
For this exercise, the continuous differentiability presupposes smooth transitions between points, ensuring no disruptions in the function鈥檚 flow.
Rolle's Theorem
Rolle's Theorem provides insights into the behavior of a differentiable function within a set interval. It states that if a function is continuous on the closed interval \( [a, b] \), differentiable on the open interval \( (a, b) \), and has equal values at the endpoints, then there exists at least one point \( c \) in between where the derivative equals zero.
Key aspects include:
  • Rolle's Theorem guarantees a "flat spot" or horizontal tangent in the interval if the conditions are met.
  • It is applicable only when \( f(a) = f(b) \).
  • The existence of such a point \( c \) indicates a potential local maximum or minimum.
In the context of this exercise, the condition \( f'(x) eq 0 \) specifies that Rolle's Theorem does not find such a point, emphasizing a single crossing of the x-axis.
Unique Root Existence
Finding a unique root within an interval where a continuous function changes sign is an intriguing result of combining the Intermediate Value Theorem with specific differential properties. The Intermediate Value Theorem ensures a zero exists due to sign changes across \[a, b\], while not allowing \( f'(x) = 0 \) prohibits additional zeros or horizontal tangents.
Consider these ideas:
  • The Intermediate Value Theorem only assures the existence of at least one root using sign change.
  • If \( f(x) \) crosses the x-axis, but \( f'(x) eq 0 \), multiple crossings cannot be supported.
  • This condition simplifies the function's graph to a single path through the axis, confirming uniqueness.
Therefore, combining these principles leads to the conclusion that the function \( f(x) \) gains only one precise root within \([a, b]\). This aligns perfectly with the mathematical assurance provided by these theorems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

a. Find a curve \(y=f(x)\) with the following properties: i) \(\frac{d^{2} y}{d x^{2}}=6 x\) ii) Its graph passes through the point (0,1) and has a horizontal tangent line there. b. How many curves like this are there? How do you know?

Sketch the graph of a differentiable function \(y=f(x)\) through the point (1,1) if \(f^{\prime}(1)=0\) and a. \(f^{\prime}(x)>0\) for \(x<1\) and \(f^{\prime}(x)<0\) for \(x>1\) b. \(f^{\prime}(x)<0\) for \(x<1\) and \(f^{\prime}(x)>0\) for \(x>1\) c. \(f^{\prime}(x)>0\) for \(x \neq 1\) d. \(f^{\prime}(x)<0\) for \(x \neq 1\)

Solve the initial value problems. $$\frac{d^{3} \theta}{d t^{3}}=0 ; \quad \theta^{\prime \prime}(0)=-2, \quad \theta^{\prime}(0)=-\frac{1}{2}, \quad \theta(0)=\sqrt{2}$$

Sketch the graph of a continuous function \(y=g(x)\) such that a. \(g(2)=2,02,\) and \(g^{\prime}(x) \rightarrow-1^{+}\) as \(x \rightarrow 2^{+}\) b. \(g(2)=2, g^{\prime}<0\) for \(x<2, g^{\prime}(x) \rightarrow-\infty\) as \(x \rightarrow 2^{-}\) \(g^{\prime}>0\) for \(x>2,\) and \(g^{\prime}(x) \rightarrow \infty\) as \(x \rightarrow 2^{+}\)

Solve the initial value problems. $$\frac{d y}{d x}=9 x^{2}-4 x+5, \quad y(-1)=0$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.