continuity

• Jan 1st 2013, 12:26 PM
AndreaM
continuity
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• Jan 1st 2013, 12:41 PM
skeeter
Re: continuity
What have done with these three problems?
• Jan 1st 2013, 12:49 PM
AndreaM
Re: continuity
How would I approach these problems? Should I graph the problems on a graphing calculator and see what kind of discontinuities the problems are?
• Jan 1st 2013, 01:24 PM
HallsofIvy
Re: continuity
The problems all ask about continuity. What do you know about "continuity"? Can you at least identify points where the function is clearly continuous and where it might not be continuous?
• Jan 2nd 2013, 10:25 AM
AndreaM
Re: continuity
I need help with this problem

Find all the constants a that make the function f(x)= 3(a-x) if x<3 continuous for all x
a2-x2 if x> or equal to 3

This is what I did on the problem so far

3(a-x) if x<3 a2-x2 if x> or equal to 3

lim-->3- 3(a-x)= limx-->3+ a2-x2

3(a-3)= 3a-9

3a-9=
limx-->3+ a2-x2

3a-9= a2-9

a2 - 3a Thats the first equation but I do not know how to get the second equation since the limits is 3 for both of the equations in the problem
• Jan 2nd 2013, 10:34 AM
AndreaM
Re: continuity
I need to find another equation so that I could plug in a^2-3a into it and find the constants for a.
• Jan 2nd 2013, 01:34 PM
skeeter
Re: continuity
Quote:

Originally Posted by AndreaM
I need to find another equation so that I could plug in a^2-3a into it and find the constants for a.

you do not need another equation.

$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x)$

$\lim_{x \to 3^-} 3(a-x) = \lim_{x \to 3^+} a^2-x^2$

$3(a-3) = a^2-9$

$3a-9 = a^2-9$

$0 = a^2-3a$

$0 = a(a-3)$

$a = 0 \, , \, a = 3$