# Continuity of a Piecewise Function

• Oct 3rd 2012, 07:09 PM
DylanRenke
[SOLVED] Continuity of a Piecewise Function

y={(x^2-4)/(x-2) if x is less than 2
{ax^2-bx+3 if x is greater than and equal to 2 and less than 3
{2x-a+b if x is greater than and equal to 3

find a and b

I have done some work and found that a=(-3/2) and b=(-7/2) but this does not work for the continuity at the point x=3

Thank you
Dylan Renke
• Oct 3rd 2012, 07:22 PM
ubhutto
Re: Continuity of a Piecewise Function
is the (a + b) in the third equation in brackets
• Oct 3rd 2012, 07:30 PM
ubhutto
Re: Continuity of a Piecewise Function
the answer for the equation is

a = 0.5
b = 0.5

plug it in and all the piecewise functions will work because all the equations will continue at x = 2
• Oct 3rd 2012, 07:45 PM
DylanRenke
Re: Continuity of a Piecewise Function
i tried that, but then when approaches 2 from the left, it equals 4, and from the right it equals 2
• Oct 3rd 2012, 07:47 PM
DylanRenke
Re: Continuity of a Piecewise Function
the a and b are not in brackets
• Oct 3rd 2012, 07:58 PM
ubhutto
Re: Continuity of a Piecewise Function
(x^2 - 4) / (x - 2) = (x+2)(x-2) / (x-2) = (x +2)
x= 2 then (2 + 2) = 4

ax^2 - bx + 3
a =0.5
b = 0.5
then (0.5 * 2^2 - 0.5 * 2) +3
1 + 3 = 4

2x - a + b
2 * 2 - 0.5 + 0.5
=4
• Oct 3rd 2012, 09:38 PM
DylanRenke
Re: Continuity of a Piecewise Function
I am just too tired, i have tried everything and just keep messing up on stupid things, the answer is 0.5 and 0.5, thank you for the help (showing work took me like an hour cause of stupid mistakes)