# Continuous

• February 9th 2013, 04:28 AM
Petrus
Continuous
Hello! This problem is almost same as last problem i posted but still i am stuck. Its problem nr 42Attachment 26919 i need help with
My progress:
(x^2-4)/(x-2)=(x+2) so at when lim goes to x->2 negative we know it shall be 4 so now we use number two that is in range 2<x<3 what we know is that when x=2 the answer shall be 4 because if its continuous there need to be lim same on negative and possitive. So we got the equation a2^2-2b=1 now im pretty stuck and dont know what to do.
• February 9th 2013, 04:48 AM
Plato
Re: Continuous
Quote:

Originally Posted by Petrus
Hello! This problem is almost same as last problem i posted but still i am stuck. Its problem nr 42Attachment 26919 i need help with

Note that $f(x) = \left\{ {\begin{array}{rr} {x + 2,} & {x \leqslant 2} \\ {ax^2 - bx + 3,} & {2 < x < 3} \\ {2x - a + b,} & {3 \leqslant x} \\ \end{array} } \right.$

$\begin{gathered} \lim _{x \to 2^ - } f(x) = 4 \hfill \\ \lim _{x \to 2^ + } f(x) = 4a - 4b + 3 \hfill \\ \lim _{x \to 3^ - } f(x) = 9a - 3b + 3 \hfill \\ \lim _{x \to 3^ + } f(x) = 6x - a + b \hfill \\ \end{gathered}$
• February 9th 2013, 04:59 AM
Petrus
Re: Continuous
Just solved it ty! Did just forget about that