continuity of piece-wise function at x = 2

[Guest]

I apologize in advance for all these questions I'm asking. I'm reviewing for the exam I have tomorrow and have lots things I find I've forgotten.

Test the continuity of each of the following functions at x = 2.

. . . . . . .. . ./ 4 - x^2, if x < 2
e) G(x) = <
. . . . . . .. . .\ 3, if x > 2

Do you do this?

. . . .4 - x^2 = 3
. . . .then replace x with 2
. . . .4 - 4 = 3
. . . .0 = 3

Then what does "0 = 3" mean? That G(x) is not continuous? Because they're not equal? Or something else?

Thank you!
________________________
Edited for clarity and formatting. --stapel

 

[wjm11]

Test the continuity of each of the following functions at x=2.

e) G(x)= 4-x^2, if x<2
3,if x greater or equal to 2

do you do this? : 4-x^2=3
then replace x with 2; 4-4=3
0=3

then umm what does 0=3 mean? not continuous? because they're not equal or ??

You have the right idea. I suggest you also think of it graphically. Draw the graph and see if the two "parts" (functions) meet at the critical point, in this case at x = 2. As you noted, they do not, so this example is discontinuous at 2.

 

[soroban]

Re: continuity

Hello, bittersweet!

Test the continuity of following function at x=2.

\(\displaystyle e)\;G(x)\;=\;\left\{\begin{array}{cc} 4\,-\,x^2 & \;\;x\,<\,2 \\3 & \;\;x\,\geq\,2\end{array}\)

Your way of writing it is confusing . . .

At \(\displaystyle x\,=\,2\), the "left" function is: \(\displaystyle \,4\,-\,2^2\:=\;0\)

At \(\displaystyle x\,=\,2\), the "right" function is: \(\displaystyle \,3\)

Since \(\displaystyle 0\,\neq\,3\), the function is discontinuous at \(\displaystyle x = 2.\)


wjm11 has the best suggestion . . . graph it.

                  |
                 ***
              *   |   *   *********
            *     |     * :
           *      |      *:
                  |       :
      - - * - - - + - - - o - - - - 
                  |       2

At \(\displaystyle x\,=\,2\), there is a "break" in the graph.
\(\displaystyle \;\;\)Therefore, it is discontinuous there.

 

[wjm11]

Soroban,

Your expertise, explanations, and graphics continue (as always) to be first class/first rate (not to belittle the contributions of many others on this forum!). Thank you for all your efforts.

wjm