/*! 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 71 Let \(f\) be differentiable at e... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(f\) be differentiable at every value of \(x\) and suppose that \(f(1)=1,\) that \(f^{\prime} < 0\) on \((-\infty, 1),\) and that \(f^{\prime} > 0\) on \((1, \infty)\) . a. Show that \(f(x) \geq 1\) for all \(x\) b. Must \(f^{\prime}(1)=0 ?\) Explain.

Short Answer

Expert verified
a. \( f(x) \geq 1 \) for all \( x \). b. Yes, \( f'(1) = 0 \).

Step by step solution

01

Understanding the Problem

The problem provides a function \( f \) that is differentiable everywhere, with specific conditions on its derivative \( f' \). It asks us to show that \( f(x) \geq 1 \) for all \( x \), and to determine if \( f'(1) = 0 \).
02

Analyzing Derivatives

The derivative \( f' \) is negative on \((-¥∞, 1)\), meaning \( f(x) \) is decreasing in this interval. Since \( f(1) = 1 \) and \( f \) is decreasing to the left of \( x = 1 \), no value of \( f(x) \) for \( x < 1 \) can exceed 1.
03

Applying Derivative Positivity

On \( (1, \infty) \), \( f' > 0 \) means that \( f(x) \) is increasing. Again, since \( f(1) = 1 \), all values \( f(x) \) for \( x > 1 \), as it increases, must remain \( \geq 1 \).
04

Conclusion for Part a

Thus, by the behavior of \( f \) as determined by \( f'(x) \), we show \( f(x) \geq 1 \) for all \( x \) since it decreases to \( x = 1 \) and then increases from \( x = 1 \).
05

Questioning Zero Derivative at 1

To determine if \( f'(1) = 0 \), we recognize that \( f \) has a local minimum at \( x = 1 \). For differentiable functions, a local minimum typically occurs where the derivative is zero, thus implying \( f'(1) = 0 \).
06

Conclusion for Part b

Therefore, \( f'(1) = 0 \) because \( f \) changes from decreasing to increasing at \( x = 1 \) indicating a local minimum where the derivative commonly is zero.

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 91Ó°ÊÓ!

Key Concepts

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

Differentiability
Differentiability is a fundamental concept in calculus that signifies whether a function has a derivative at each point in its domain. A function is differentiable at a point if it is smooth or continuous there, without any abrupt changes in direction or sharp corners. When a function is differentiable across its entire domain, it means you can compute how fast it is changing at any point within that domain.
To determine if a function is differentiable, check its piecewise structure or look for points where the behavior alters.
  • If a function is differentiable at a point, it means the derivative exists and is finite at that point.
  • A differentiable function is necessarily continuous, but the reverse isn't always true; not all continuous functions are differentiable.
For example, the function given in our exercise, denoted as \( f \), is stated to be differentiable everywhere, enabling us to explore how the function behaves around specific points, such as where the derivative changes from negative to positive.
Derivative
The concept of a derivative in calculus describes the rate at which a function is changing at any given point. It can be seen as a measure of the slope of the function at a particular point. The derivative, often denoted as \( f' \) or \( \frac{df}{dx} \), gives critical information about the function's instantaneous rate of change.
  • If \( f'(x) > 0 \), the function \( f \) is increasing at \( x \).
  • If \( f'(x) < 0 \), the function \( f \) is decreasing at \( x \).
  • When \( f'(x) = 0 \), the function might have a local maximum, a local minimum, or a stationary point at \( x \).
In our exercise, the derivative of \( f \), \( f' \), is negative on the interval \((-\infty, 1)\) and positive on \((1, \infty)\). This information helps us determine the function's behavior, indicating it decreases towards \( x=1 \) and increases thereafter, forming a typical peak or valley scenario.
Decreasing and Increasing Functions
In calculus, identifying increasing and decreasing functions is crucial to understanding their overall behavior. An increasing function is one where the value of \( f(x) \) gets larger as \( x \) gets larger, and conversely, a decreasing function sees \( f(x) \) getting smaller as \( x \) increases. Derivatives help here by indicating this behavior based on their sign.
For example, let's consider the following:
  • If \( f' < 0 \) on an interval, \( f \) is decreasing on that interval.
  • If \( f' > 0 \) on an interval, \( f \) is increasing on that interval.
The function \( f \) in our problem is decreasing for all \( x \) less than 1 due to negatively-valued derivatives, and it's increasing for all \( x \) greater than 1 where the derivatives are positive. This behavior around \( x = 1 \) suggests the presence of a local minimum at that point, proving key for solving specific parts of the problem, such as establishing that \( f(x) \geq 1 \) everywhere by recognizing this minimum.

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

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=\sin x \cos x, \quad 0 \leq x \leq \pi$$

Identify the inflection points and local maxima and minima of the functions graphed. Identify the intervals on which the functions are concave up and concave down. $$y=\tan x-4 x,-\frac{\pi}{2}< x<\frac{\pi}{2}$$

Cubic curves What can you say about the inflection points of a cubic curve \(y=a x^{3}+b x^{2}+c x+d, a \neq 0 ?\) Give reasons for your answer.

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x^{2 / 3}\left(\frac{5}{2}-x\right)$$

Distance between two ships At noon, ship \(A\) was 12 nautical miles due north of ship \(B .\) Ship \(A\) was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship \(B\) was sailing east at 8 knots and continued to do so all day. \begin{equation} \begin{array}{l}{\text { a. Start counting time with } t=0 \text { at noon and express the dis- }} \\ \quad {\text { tance } s \text { between the ships as a function of } t \text { . }} \\ {\text { b. How rapidly was the distance between the ships changing at }} \\ \quad {\text { noon? One hour later? }} \\\ {\text { c. The visibility that day was } 5 \text { nautical miles. Did the ships }} \\ \quad {\text { ever sight each other? }} \\ {\text { d. Graph } s \text { and } d s / d t \text { together as functions of } t \text { for }-1 \leq t \leq 3} \\ \quad {\text { using different colors if possible. Compare the graphs and }} \\ \quad {\text { reconcile what you see with your answers in parts (b) }} \\ \quad {\text { and (c). }} \\ {\text { e. The graph of } d s / d t \text { looks as if it might have a horizontal }} \\\ \quad {\text { asymptote in the first quadrant. This in turn suggests that }} \\\ \quad {d s / d t \text { approaches a limiting value as } t \rightarrow \infty . \text { What is this }} \\ \quad {\text { value? What is its relation to the ships' individual speeds? }} \end{array} \end{equation}

See all solutions

Recommended explanations on Math Textbooks

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.