text: # Continuity
  
  Printable View
  
  • Feb 3rd 2011, 12:18 PM
  bodel
  Continuity
  Hey everyone, I can't figure this problem out:
  
  Find the constants a and b that will make the following function continuous everywhere.
  
  f(x)=
  • $a cosx$ if x<0
  • sin $\left(\frac{pi}{2}x+3)$ if 0<=x<=2
  • $x^2-x+b$ if x>2
  Thanks in advance!
  • Feb 3rd 2011, 12:39 PM
  skeeter
  Quote:
  
  Originally Posted by bodel
  Hey everyone, I can't figure this problem out:
  
  Find the constants a and b that will make the following function continuous everywhere.
  
  f(x)=
  • $a cosx$ if x<0
  • sin $\left(\frac{pi}{2}x+3)$ if 0<=x<=2
  • $x^2-x+b$ if x>2
  Thanks in advance!
  
  I assume you are aware of the definition of continuity at a point ...
  
  $\displaystyle \lim_{x \to 0^+} f(x) = \sin(3)
  $
  
  $f(0) = \sin(3)$
  
  therefore $\displaystyle \lim_{x \to 0^-} f(x) = \sin(3) = a\cos(0)$
  
  $a = ?$
  
  $\displaystyle \lim_{x \to 2^-} f(x) = \sin(\pi+3) = -\sin(3)$
  
  $f(2) = -\sin(3)$
  
  therefore, $\displaystyle \lim_{x \to 2^+} f(x) = 2^2 - 2 + b = -\sin(3)$
  
  $b = ?$
  • Feb 3rd 2011, 12:59 PM
  bodel
  Thanks skeeter!! I actually got the same answers for both sides of the limit of x=0 and x=2. I was getting confused because I knew they had to equal each other, but I didn't actually set the answers equal and solve for the constant.
  
  Would a = sin(3) and b = -sin(3)-2
  • Feb 3rd 2011, 01:22 PM
  Plato
  Quote:
  
  Originally Posted by bodel
  I actually got the same answers for both sides of the limit of x=0 and x=2. Would a = sin(3) and b = -sin(3)-2
  
  The the value of a is correct.
  
  $b=-\sin(3)-2.$ WHY?
  • Feb 3rd 2011, 02:01 PM
  bodel
  $\displaystyle \lim_{x \to 2^+} f(x) = 2^2 - 2 + b = -\sin(3)$
  
  $b = ?$
  
  1. 4-2+b = -sin(3)
  2. 2+b = -sin(3)
  3. b = -sin(3) - 2?
  
  I'm prone to dumb mistakes so please point them out if you find some.
  Quote:
  
  Originally Posted by Plato
  The the value of a is correct.
  
  $b=-\sin(3)-2.$ WHY?
  
  • Feb 3rd 2011, 02:22 PM
  skeeter
  $b = -\sin(3) - 2$
  
  works for me.
explanation: The document is a math forum post where a user named bodel asks for help in finding the constants \(a\) and \(b\) to make a piecewise function continuous everywhere. Another user, skeeter, provides a detailed solution, and bodel confirms the solution. The post contains a question and an answer.
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  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Continuity
  
  Printable View
  
  • Feb 3rd 2011, 12:18 PM
  bodel
  Continuity
  Hey everyone, I can't figure this problem out:
  
  Find the constants a and b that will make the following function continuous everywhere.
  
  f(x)=
  • $a cosx$ if x<0
  • sin $\left(\frac{pi}{2}x+3)$ if 0<=x<=2
  • $x^2-x+b$ if x>2
  Thanks in advance!
  • Feb 3rd 2011, 12:39 PM
  skeeter
  Quote:
  
  Originally Posted by bodel
  Hey everyone, I can't figure this problem out:
  
  Find the constants a and b that will make the following function continuous everywhere.
  
  f(x)=
  • $a cosx$ if x<0
  • sin $\left(\frac{pi}{2}x+3)$ if 0<=x<=2
  • $x^2-x+b$ if x>2
  Thanks in advance!
  
  I assume you are aware of the definition of continuity at a point ...
  
  $\displaystyle \lim_{x \to 0^+} f(x) = \sin(3)
  $
  
  $f(0) = \sin(3)$
  
  therefore $\displaystyle \lim_{x \to 0^-} f(x) = \sin(3) = a\cos(0)$
  
  $a = ?$
  
  $\displaystyle \lim_{x \to 2^-} f(x) = \sin(\pi+3) = -\sin(3)$
  
  $f(2) = -\sin(3)$
  
  therefore, $\displaystyle \lim_{x \to 2^+} f(x) = 2^2 - 2 + b = -\sin(3)$
  
  $b = ?$
  • Feb 3rd 2011, 12:59 PM
  bodel
  Thanks skeeter!! I actually got the same answers for both sides of the limit of x=0 and x=2. I was getting confused because I knew they had to equal each other, but I didn't actually set the answers equal and solve for the constant.
  
  Would a = sin(3) and b = -sin(3)-2
  • Feb 3rd 2011, 01:22 PM
  Plato
  Quote:
  
  Originally Posted by bodel
  I actually got the same answers for both sides of the limit of x=0 and x=2. Would a = sin(3) and b = -sin(3)-2
  
  The the value of a is correct.
  
  $b=-\sin(3)-2.$ WHY?
  • Feb 3rd 2011, 02:01 PM
  bodel
  $\displaystyle \lim_{x \to 2^+} f(x) = 2^2 - 2 + b = -\sin(3)$
  
  $b = ?$
  
  1. 4-2+b = -sin(3)
  2. 2+b = -sin(3)
  3. b = -sin(3) - 2?
  
  I'm prone to dumb mistakes so please point them out if you find some.
  Quote:
  
  Originally Posted by Plato
  The the value of a is correct.
  
  $b=-\sin(3)-2.$ WHY?
  
  • Feb 3rd 2011, 02:22 PM
  skeeter
  $b = -\sin(3) - 2$
  
  works for me.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a math forum post where a user named bodel asks for help in finding the constants \(a\) and \(b\) to make a piecewise function continuous everywhere. Another user, skeeter, provides a detailed solution, and bodel confirms the solution. The post contains a question and an answer.
  
  Label: question-and-answer