Thread: Continuity

The function Abs[x - 2]/(x^2 - 5x + 6) is not defined for x=2. Is there a value that could be given for f(2) that would make the function continuous at 2?
Give reasons for your answer.

You dropped an x on the denominator I think. Try factoring.

Originally Posted by maddas

You dropped an x on the denominator I think. Try factoring.

yea my bad. You factorise it and simplify to get 1/(x-3) but then what would that mean?

Are you sure that's how it reduces? Make sure to consider both x>2 and x<2....

Originally Posted by maddas

Are you sure that's how it reduces? Make sure to consider both x>2 and x<2....

|x-2|/(x-3)(x-2) = 1/(x-3)
or cant i cancel them cause its abs?

You tell me. Are both sides of that equation actually equal for all x? What about x=1?

Because is...

(1)

... is...



(2)

... so that f(x) isn't continous for no matter which is the value You assign to it in that point...

Kind regards

^
Thank you

Originally Posted by maddas

You tell me. Are both sides of that equation actually equal for all x? What about x=1?

time waster...

Then they're not equal. Anyway, chisigma posted a solution now.

edit: yea, cause its such a waste of time to know that 1/2 is not -1/2.

Maddas' point when he wrote "You tell me. Are both sides of that equation actually equal for all x?" was that you cannot cancel "x- 2" because at x= 2, that would be dividing by 0.

[tex]\frac{(x- 3)(x- 2)}{x- 2}= x- 3[/math for all x except x= 2. That's why the original function, is not defined at x= 2.

Fortunately, the definition of " " says that "Given , there exist such that if 0< , then ". That "0< |x- a|" (which many people forget to write) means that the value of f(x) at x= a has no effect whatsoever on the lim as x goes to a.

Although [tex]\frac{(x- 3)(x- 2)}{x- 2}[tex] is not defined at x= 2, it has a limit there and you can make the function continuous by defining f(2) to be that limit.