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- Jan 1st 2013, 12:26 PMAndreaMcontinuity
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- Jan 1st 2013, 12:41 PMskeeterRe: continuity
What have done with these three problems?

- Jan 1st 2013, 12:49 PMAndreaMRe: 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 PMHallsofIvyRe: 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 AMAndreaMRe: 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

a^{2}-x^{2}if x> or equal to 3

This is what I did on the problem so far

3(a-x) if x<3 a^{2}-x^{2}if x> or equal to 3

lim-->3- 3(a-x)= lim_{x-->3+}a^{2}-x^{2 3(a-3)= 3a-9 3a-9=}lim_{x-->3+}a^{2}-x^{2 }3a-9= a^{2}-9

a^{2 }- 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 AMAndreaMRe: 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 PMskeeterRe: continuity
you do not need another equation.

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

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

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

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

$\displaystyle 0 = a^2-3a$

$\displaystyle 0 = a(a-3)$

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