/*! 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 5 Let $$f(x)=\left\\{\begin{arra... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let $$f(x)=\left\\{\begin{array}{ll}{\frac{1}{1-x},} & {0 \leq x<1} \\ {0,} & {x=1}\end{array}\right.$$ Explain why \(f\) has a minimum value but no maximum value on the closed interval \([0,1] .\)

Short Answer

Expert verified
The minimum value is 0 at \( x = 1 \), with no maximum due to \( f(x) \to \infty \) as \( x \to 1^- \).

Step by step solution

01

Understanding the Domain of f(x)

The function \( f(x) \) is defined as \( \frac{1}{1-x} \) for \( 0 \leq x < 1 \) and \( 0 \) when \( x = 1 \). This means \( f(x) \) is continuous on the interval \([0,1)\) and has a defined value at \( x = 1 \).
02

Evaluating f(x) at Specific Points

Let's evaluate \( f(x) \) at the interval boundaries: \( f(0) = \frac{1}{1-0} = 1 \). At \( x = 1 \), \( f(1) = 0 \).
03

Analyzing the Behavior of f(x) as x Approaches 1

As \( x \) approaches 1 from the left (i.e., \( x \to 1^- \)), the function \( f(x) = \frac{1}{1-x} \) increases without bound, meaning \( \lim_{{x \to 1^-}} f(x) = +\infty \). This indicates \( f(x) \) has no upper bound as \( x \to 1 \).
04

Determining the Minimum Value of f(x)

Since \( f(0) = 1 \) and \( f(1) = 0 \), the minimum value of \( f(x) \) on \([0,1]\) is \( 0 \), achieved at \( x = 1 \).
05

Concluding No Maximum Value Exists

Because \( f(x) \) tends to infinity as \( x \to 1^- \), there is no maximum value of \( f(x) \) on the interval \([0,1]\). Thus, while there is a minimum value, the function does not attain a maximum value on this interval.

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.

Piecewise functions
Piecewise functions are a type of function made up of multiple sub-functions, each of which applies to a specific interval or piece of the domain. This means that instead of one formula defining the function across its entire domain, a piecewise function has different formulas for different sections of its domain. In our example, the function is defined as:
  • \( f(x) = \frac{1}{1-x} \) for \( 0 \leq x < 1 \)
  • \( f(x) = 0 \) when \( x = 1 \)
This means that for any \( x \) less than 1, we use the first formula, while exactly at \( x = 1 \), we use the second formula where \( f(x) \) is 0. Piecewise functions are helpful in modeling situations where a rule changes at a boundary, which is common in real-world problems.
Continuity
In calculus, continuity of a function at a point means there are no breaks, jumps, or holes in the graph of the function at that point. For a function to be continuous on an interval, it must be continuous at every point within that interval. In our function, \( f(x) \) is continuous on the open interval \([0, 1)\). It is not continuous at \( x = 1 \). This is because the value of the function jumps from what it would be as it approaches 1 from the left, to 0 exactly at 1.
To check continuity mathematically, we see if the limit of the function as \( x \) approaches a point equals the value of the function at that point. Since as \( x \to 1^- \), \( f(x) \to +\infty \), but \( f(1) = 0 \), the function is not continuous at \( x = 1 \). This discontinuity at the boundary cumulatively affects the closed interval \([0, 1]\), making it non-continuous.
Limits
Limits are fundamental in calculus as they help describe the behavior of functions as they approach specific points. The limit tells us what value a function is approaching as the input gets closer to a certain point, but not necessarily what the value is at that point. For \( f(x) \), the interesting limit is as \( x \) approaches 1 from the left, written as \( \lim_{{x \to 1^-}} f(x) \).
In the example, this limit is +\infty, meaning as \( x \) gets closer and closer to 1, \( f(x) \) grows larger without bound. Still, this is different from the value of the function at 1, which is defined to be 0. So while the limit gives us an idea of the behavior surrounding the point, it does not necessarily equate to the function's value at that point, showcasing a critical point about limits and continuity.
Extrema
Extrema refer to the maximum and minimum values of a function. Every continuous function defined on a closed interval should have a minimum and a maximum value due to the Extreme Value Theorem. However, our function only satisfies part of this condition.
  • The minimum value of \( f(x) \) on the interval \([0,1]\) is 0, occurring at \( x = 1 \).
  • There is no maximum value because as \( x \to 1^- \), \( f(x) \to +\infty \). The function increases without bound as \( x \) approaches 1 from the left.
Therefore, the function has a lower bound but no upper bound within this domain. This happens because the function is not actually attaining \(+\infty\), instead it just perpetually increases. Understanding extrema in piecewise functions can thus be a bit more complex compared to straightforward continuous functions on closed intervals.

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

An annuity is a sequence of equal payments that are paid or received at regular time intervals. For example, you may want to deposit equal amounts at the end of each year into an interest-bearing account for the purpose of accumulating a lump sum at some future time. If, at the end of each year, interest of \(i \times 100 \%\) on the account balance for that year is added to the account, then the account is said to pay \(i \times 100 \%\) interest, compounded annually. It can be shown that if payments of \(Q\) dollars are deposited at the end ofeach year into an account that pays \(i \times 100 \%\) compounded annually, then at the time when the \(n\) th payment and the accrued interest for the past year are deposited, the amount \(S(n)\) in the account is given by the formula $$S(n)=\frac{Q}{i}\left[(1+i)^{n}-1\right]$$ Suppose that you can invest \(\$ 5000\) in an interest-bearing account at the end of each year, and your objective is to have \(\$ 250,000\) on the 25th payment. Approximately what annual compound interest rate must the account pay for you to achieve your goal? [Hint: Show that the interest rate \(i\) satisfies the equation \(50 i=(1+i)^{25}-1,\) and solve it using Newton's Method.]

Verify that the hypotheses of Rolle's Theorem are satisfied on the given interval, and find all values of \(c\) in that interval that satisfy the conclusion of the theorem. $$ f(x)=\cos x ;[\pi / 2,3 \pi / 2] $$

Let \(s_{A}=15 t^{2}+10 t+20\) and \(s_{B}=5 t^{2}+40 t, t \geq 0,\) be the position functions of cars \(A\) and \(B\) that are moving along parallel straight lanes of a highway. (a) How far is car \(A\) ahead of car \(B\) when \(t=0\) ? (b) At what instants of time are the cars next to each other? (c) At what instant of time do they have the same velocity? Which car is ahead at this instant?

In each part, explain what happens if you apply Newton's Method to a function \(f\) when the given condition is satisfied for some value of \(n .\) (a) \(f\left(x_{n}\right)=0\) (b) \(x_{n+1}=x_{n}\) (c) \(x_{n+2}=x_{n} \neq x_{n+1}\)

If \(p(x)\) is a polynomial, discuss the usefulness of knowing zeros for \(p, p^{\prime},\) and \(p^{\prime \prime}\) when determining information about the graph of \(p .\)

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.