Making a piecewise function continuous

  • #1

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

 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi meaganjulie! Welcome to PF! :wink:
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:
 
  • #3
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 ive got.
 
  • #4
Char. Limit
Gold Member
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14
Just figure out what c and b values make the first and second equal at x=0.

The other one follows a similar process.
 
  • #5
i dont understand how to find the values for c and b because i dont 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?
 
  • #6
tiny-tim
Science Advisor
Homework Helper
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Hi meaganjulie! :smile:

(just got up :zzz: …)
i dont understand how to find the values for c and b because i dont 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.
 

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