Can someone help with this queation

[Jared123]

If possible, find all values of ? for which the following function is continuous for all real values of ?:

?? + 3? 2 , ? < 2
?(?) = {
6? − ?? 2 , ? > 2 .

 

[Otis]

Hi Jared. What have you learned about continuous graphs?

Here's a question for you to consider: Does the domain of f include 2?

?

 

[HallsofIvy]

A function, f(x), is "continuous at x= a" if and only if
1) f(a) exists
2) \(\displaystyle \lim_{x\to a} f(x)\) exists
3) \(\displaystyle \lim_{x\to a} f(x)= f(a)\)

It should be obvious that this function is continuous for all x except possibly x= 2.

So what are f(2) and \(\displaystyle \lim_{x\to 2} f(x)\)? What value of a makes those the same?

(Unfortunately, the way you have this written, f(2) does not exist for ANY value of a! I suspect that one of those "<" or ">" should be "\(\displaystyle \le\)" or "\(\displaystyle \ge\)".)

 

[pka]

If possible, find all values of ? for which the following function is continuous for all real values of ?:
?? + 3? 2 , ? < 2
?(?) = {
6? − ?? 2 , ? > 2 .

Let's make the function readable: \(f(x)=ax+3x^2,~x\le 2\text{ and }6x-ax^2,~x>2\)
We need to make \(\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} f(x)\)
where \(\mathop {\lim }\limits_{x \to {2^ - }} f(x) =2a+12\text{ and } \mathop {\lim }\limits_{x \to {2^ + }} f(x)=12-4a\)
Can you find an \(a\) making those equal?

 

[HallsofIvy]

Thank you. But how did you know that the first one was \(\displaystyle x\le 2\) rather than x< 2 as was given?

 

[Steven G]

Exactly! A necessary condition for f(x) to be continuous at x=2 is that f(x) must be defined at x=2. In layman language, a function is continuous at x=2 if you can draw the function from some point to the left of x =2 to some point to the right of x=2 without lifting your pencil. How can you do that if the function is not defined at x=2!

More rigorously you can NOT satisfy condition #3 posted by HallsofIvy

 

[pka]

Thank you. But how did you know that the first one was \(\displaystyle x\le 2\) rather than x< 2 as was given?

Oh, come one George it makes no difference for continuous functions. DOES IT???

 

[Steven G]

Oh, come one George it makes no difference for continuous functions. DOES IT???

Maybe, but for discontinuous functions it certainly does.

It is not a bad exercise or even test problem to ask that same exact question where there is no equality at x=2 at all. The function is not continuous at x=2.

 

[HallsofIvy]

Yes, where the "=" is does not matter for this question. And I just noticed that the title says "queation". A mélange of "question" and "equation"! I love it!

 

[Steven G]

Yes, where the "=" is does not matter for this question. And I just noticed that the title says "queation". A mélange of "question" and "equation"! I love it!

I am saying that there is no equality sign for this problem as the OP states. No what what value one picks for a, f(x) will not be continuous at x=2. Removable discontinuity, yes, continuous, no!