# Help with continuous function.

• Sep 9th 2008, 11:13 AM
Afterme
Help with continuous function.
Hello guys,
I'm trying to understand this problem but having a hard time. Any help to point me in the right direction will be appreciated thanks (Rock)

For what value(s) of the constant c is the following function continuous on (−∞,∞)?

f(x) = cx+1, x (less then or equal) 2 and cx^2, x > 2

Sorry about the format. It's supposed to be set up to look like a piecewise function but i can't seem to get it like that
• Sep 9th 2008, 11:32 AM
Shyam
Quote:

Originally Posted by Afterme
Hello guys,
I'm trying to understand this problem but having a hard time. Any help to point me in the right direction will be appreciated thanks (Rock)

For what value(s) of the constant c is the following function continuous on (−∞,∞)?

f(x) = cx+1, x (less then or equal) 2 and cx^2, x > 2

Sorry about the format. It's supposed to be set up to look like a piecewise function but i can't seem to get it like that

$\displaystyle f\left( x \right) = \left\{ \begin{gathered} cx + 1,{\text{ }}x \leqslant 2 \hfill \\ cx^2 {\text{ }}\;\;\;\;\;x > 2 \hfill \\ \end{gathered} \right. \hfill \\$

$\displaystyle {\text{For }}f\left( x \right){\text{ to be continuous in }}\left( { - \infty ,\infty } \right){\text{, it should be continuous at }}x = 2. \hfill \\$

$\displaystyle {\text{Left hand limit,}} = \mathop {\lim }\limits_{x \to 2 - } f\left( x \right) = \mathop {\lim }\limits_{x \to 2 - } \left( {cx + 1} \right) = 2c + 1 \hfill \\$

$\displaystyle {\text{Right hand limit,}} = \mathop {\lim }\limits_{x \to 2 + } f\left( x \right) = \mathop {\lim }\limits_{x \to 2 + } \left( {cx^2 } \right) = 4c \hfill \\$

$\displaystyle f\left( 2 \right) = 2c + 1 \hfill \\$

$\displaystyle {\text{For }}f\left( x \right){\text{ to be continuous at }}x = 2,{\text{ }} \Rightarrow \mathop {\lim }\limits_{x \to 2 - } f\left( x \right) = \mathop {\lim }\limits_{x \to 2 + } f\left( x \right) = f\left( 2 \right) \hfill \\$

$\displaystyle \Rightarrow 2c + 1 = 4c \hfill \\$

$\displaystyle \Rightarrow c = \frac{1} {2} \hfill \\$
• Sep 9th 2008, 03:12 PM
Afterme
You say for F(x) to be continuous, it has to be continuous at X = 2. Why though. I'm a bit confused on that part.
• Sep 9th 2008, 03:25 PM
Plato
Quote:

Originally Posted by Afterme
You say for F(x) to be continuous, it has to be continuous at X = 2. Why though.

Do you know what it means for a function on the real numbers is continuous, period?
• Sep 9th 2008, 04:29 PM
fogel1497
For a function to be continuous the "Right hand limit" and "Left hand limit" must be approaching the same number for any give number.

Both of those functions in the piecewise are continuous from negative infiniti to positive infiniti. But since at x=2 where the piecewise function 'switches functions' the two functions are not approaching the same number from the left and the right.

So you must plug in a value for 'c' that will make them approach the same number. You do this by setting the two equations equal to each other, so at that point both parts of the piecewise equal the same number, and are therefore, (in this case at least), approaching the same number.
• Sep 9th 2008, 04:44 PM
Afterme
Thanks fogel that cleared it up. (Rock)