# Continuity

• 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)=
• $\displaystyle a cosx$ if x<0
• sin$\displaystyle \left(\frac{pi}{2}x+3)$ if 0<=x<=2
• $\displaystyle x^2-x+b$ if x>2
• 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)=
• $\displaystyle a cosx$ if x<0
• sin$\displaystyle \left(\frac{pi}{2}x+3)$ if 0<=x<=2
• $\displaystyle x^2-x+b$ if x>2

I assume you are aware of the definition of continuity at a point ...

$\displaystyle \displaystyle \lim_{x \to 0^+} f(x) = \sin(3)$

$\displaystyle f(0) = \sin(3)$

therefore $\displaystyle \displaystyle \lim_{x \to 0^-} f(x) = \sin(3) = a\cos(0)$

$\displaystyle a = ?$

$\displaystyle \displaystyle \lim_{x \to 2^-} f(x) = \sin(\pi+3) = -\sin(3)$

$\displaystyle f(2) = -\sin(3)$

therefore, $\displaystyle \displaystyle \lim_{x \to 2^+} f(x) = 2^2 - 2 + b = -\sin(3)$

$\displaystyle b = ?$
• 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.

$\displaystyle b=-\sin(3)-2.$ WHY?
• Feb 3rd 2011, 02:01 PM
bodel
$\displaystyle \displaystyle \lim_{x \to 2^+} f(x) = 2^2 - 2 + b = -\sin(3)$

$\displaystyle b = ?$

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.

$\displaystyle b=-\sin(3)-2.$ WHY?

• Feb 3rd 2011, 02:22 PM
skeeter
$\displaystyle b = -\sin(3) - 2$

works for me.