# Continuity

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• 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.
The text passage contains a mathematical question related to finding constants to ensure a piecewise function is continuous everywhere. The question is answered in the text, with the values of the constants being derived and confirmed.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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.
user
The text passage contains a mathematical question related to finding constants to ensure a piecewise function is continuous everywhere. The question is answered in the text, with the values of the constants being derived and confirmed.

TYPE: SINGLE_QUESTION_WITH_ANSWER