Making a piecewise function continuous

Click For Summary

Homework Help Overview

The discussion revolves around determining the values of b and c that ensure the continuity of a piecewise function defined on the entire real line. The function consists of three segments, each defined for different intervals of x, and participants are exploring the limits at specific points to achieve continuity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are examining the limits of the function at critical points, particularly x=0 and x=2, to find conditions on b and c. There is a focus on ensuring that the left-hand and right-hand limits match at these points.

Discussion Status

The conversation includes various interpretations of the limits involved. Some participants have identified the limit as x approaches 0, while others are questioning how to apply this to find the necessary values for b and c. There is an exchange of ideas about the relationships between the segments of the piecewise function.

Contextual Notes

Participants note that the first segment of the function does not involve b or c, which raises questions about how to utilize the limits from the other segments to establish continuity. There is also mention of needing to match limits at x=2, but uncertainty remains regarding how to determine those limits.

meaganjulie
Messages
3
Reaction score
0

Homework Statement



find the values of b and c that make the function f continuous on (-[tex]\infty[/tex],[tex]\infty[/tex])

f(x) = [tex]\frac{sin2x}{x}[/tex] if x< 0
3-3c+b(x+1) if 0[tex]\leq[/tex]x<2
5-cx+bx^2 if x[tex]\geq[/tex] 2

Homework Equations



lim as x [tex]\rightarrow[/tex] 0- of [tex]\frac{sin2x}{x}[/tex]
works out to be 0

The Attempt at a Solution

 
Physics news on Phys.org
Welcome to PF!

Hi meaganjulie! Welcome to PF! :wink:
meaganjulie said:
lim as x [tex]\rightarrow[/tex] 0- of [tex]\frac{sin2x}{x}[/tex]
works out to be 0

Nope :redface:

what makes you think that? :smile:
 
nevermind, that limit is actually 2. the limit for zero from the right also must equal 2, and the right and left limits at x=2 must also match.

thats all I've got.
 
Just figure out what c and b values make the first and second equal at x=0.

The other one follows a similar process.
 
i don't understand how to find the values for c and b because i don't understand how i can use the point x=0 because the first part of the function has no c or b values. i have to use the last to equations at x=2, but how do i know what that limit is?
 
Hi meaganjulie! :smile:

(just got up :zzz: …)
meaganjulie said:
i don't understand how to find the values for c and b because i don't understand how i can use the point x=0 because the first part of the function has no c or b values. i have to use the last to equations at x=2, but how do i know what that limit is?

The first equation tells you that the limit at x = 0 must be 2.

The second equation tells you conditions on b and c which agree with that limit (at x = 0), and that gives you a formula (in b and c) for the limit at x = 2.

And the third equation tells you conditions on b and c which agree with that limit at x = 2.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K