/*! 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 33 The number of points on \([0,2]\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 33

The number of points on \([0,2]\) where $f(x)= \begin{cases}x\\{x\\}+1 & 0 \leq x<1 \\ 2-\\{x\\} & 1 \leq x \leq 2\end{cases}$ fails to be continuous or derivable is (A) 0 (B) 1 (C) 2 (D) 3

Short Answer

Expert verified
Answer: (C) 2

Step by step solution

01

Check for continuity at x=1

First, we need to check if the limit of the function exists at x=1. The limit from the left side is equal to the limit of the first piece at x=1: \(\lim\limits_{x \to 1^-} (x + 1) = 1+1 = 2\) And the limit from the right side is equal to the limit of the second piece at x=1: \(\lim\limits_{x \to 1^+} (2 - \{x\}) = 2 - \{1\} = 1\) Since the left and right limits are different, the function is discontinuous at x=1. Thus, the function fails to be continuous at 1 point.
02

Check for derivability at x=1

To check for derivability, we need to find the first derivative of the function for each piece and check if they are equal at x=1. First, find the derivative of the first piece: \(f'(x) = \frac{d}{dx} (x + 1) = 1\) For the second piece, we have: \(f'(x) = \frac{d}{dx} (2 - \{x\}) = 0\) Clearly, the first derivative for both pieces is not equal at x=1. Thus, the function fails to be derivable at 1 point.
03

Count the number of points

Now that we have found out that the function fails to be continuous and derivable at x=1, we can conclude that the number of points where the function is either discontinuous or non-differentiable is: 1 (discontinuity) + 1 (non-derivability) = 2 Therefore, the correct answer is (C) 2.

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!

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

Let \(\mathrm{h}(\mathrm{x})\) be differentiable for \(\mathrm{all} \mathrm{x}\) and let $\mathrm{f}(\mathrm{x})=\left(\mathrm{k} \mathrm{x}+\mathrm{e}^{x}\right)$ \(\mathrm{h}(\mathrm{x})\) where \(\mathrm{k}\) is some constant. If $h(0)=5, \mathrm{~h}^{\prime}(0)=-2\( and \)f^{\prime}(0)=18\( then the value of \)k$ is equal to (A) 5 (B) 4 (C) 3 (D) \(2.2\)

If \(f(x) \cdot f(y)=f(x)+f(y)+f(x y)-2 \forall x, y \in R\) and if \(f(x)\) is not a constant function, then the value of \(f(1)\) is (A) 1 (B) 2 (C) 0 (D) \(-1\)

Let \(\mathrm{f}(\mathrm{x})\) be differentiable at \(\mathrm{x}=\mathrm{h}\) then \(\lim _{x \rightarrow h} \frac{(x+h) f(x)-2 h f(h)}{x-h}\) is equal to (A) \(f(h)+2 h f^{\prime}(h)\) (B) \(2 \mathrm{f}(\mathrm{h})+\mathrm{hf}^{\prime}(\mathrm{h})\) (C) \(\mathrm{hf}(\mathrm{h})+2 \mathrm{f}^{\prime}(\mathrm{h})\) (D) \(h f(h)-2 f^{\prime}(h)\)

If $f(x)=\left\\{\begin{array}{ll}{[x]+\sqrt{\\{x\\}}} & x<1 \\\ \frac{1}{[x]+\\{x\\}^{2}} & x \geq 1\end{array}\right.$, then [where [. ] and \\{ - \(\\}\) represent greatest integer and fractional part functions respectively] (A) \(\mathrm{f}(\mathrm{x})=\) is continuous at \(\mathrm{x}=1\) but not differentiable (B) \(\mathrm{f}(\mathrm{x})\) is not continuous at \(\mathrm{x}=1\)

I.et \(\mathrm{f}(\mathrm{x})\) be a function such that \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})\) and \(f(x)=\sin x g(x)\) for all \(x, y \in R\), If \(g(x)\) is a continuous function such that \(\mathrm{g}(0)=\mathrm{K}\), then \(f^{\prime}(\mathrm{x})\) is equal to (A) \(\mathrm{K}\) (B) \(\mathrm{Kx}\) (C) \(\mathrm{Kg}(\mathrm{x})\) (D) none
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

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.