# Piece wise function help

• February 2nd 2013, 08:24 PM
Oldspice1212
Piece wise function help
f(x) =
 x2 − 4 x − 2
if x < 2
ax2bx + 3 if 2x < 3
4xa + b if x ≥ 3

OK guys so I'm having trouble with this, it includes a lot of algebra steps but I'll put some of them to show that I DID attempt it several times, not sure if I'm just tired or I messed up at some point....

I factored out at the start of coarse...
(x-2)(x+2)/ (x-2) = ax^2 - bx + 3 -----> (2+2) = a(2)^2-b(2)+3
4 = 4a - 2b +3
4 - 3 = 4a - 2b
1 = 4a - 2b
1+2b/4 = 4a/4
1/4+1/2b = a

ax^2 -bx + 3 = 4x - ab
a(3)^2 - b(3) +3 = 4(3) - a + b
9a + a - 3b - b = 12- 3
10a - 4b = 9
10(1/4 + 1/2b) - 4b = 9
5/2 + 5b - 4b = 9
5/2 - 1b = 9
-1b = 13/2
= -13/2

1/4 + 1/2(-13/2) = a
1/4 + (-13/4) = a
- 3 = a

a = -3 b = -13/2
• February 2nd 2013, 08:51 PM
Prove It
Re: Piece wise function help
This is extremely hard to read. Is your function \displaystyle \begin{align*} f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} \textrm{ if } x < 2 \\ a x^2 - b x + 3 \textrm{ if } 2 \leq x < 3 \\ 4x - a + b \textrm{ if } x \geq 3 \end{cases} \end{align*}?
• February 2nd 2013, 09:43 PM
Oldspice1212
Re: Piece wise function help
Yes exactly right, sorry about that, I wasn't sure how to make it look like that so I copied the question exactly :S
• February 2nd 2013, 09:46 PM
Prove It
Re: Piece wise function help
What are you actually trying to do with this question? Are you trying to find the values of a and b which make this function continuous?
• February 2nd 2013, 10:05 PM
Oldspice1212
Re: Piece wise function help
Quote:

Originally Posted by Prove It
What are you actually trying to do with this question? Are you trying to find the values of a and b which make this function continuous?

Yes sir, sorry for not mentioning that been up for longer then I should.
• February 2nd 2013, 11:01 PM
Prove It
Re: Piece wise function help
It should be clear that the function needs to approach 4 if you make x approach 2 from the left (why). For the function to be continuous at x = 2, it needs to approach the same value from the right. So that means \displaystyle \begin{align*} a(2)^2 - b(2) + 3 = 2 \end{align*}.

For the function to be continuous at x = 3, we require the function to approach the same value from the left as from the right. So \displaystyle \begin{align*} a(3)^2 - b(3) + 3 = 4(3) - a + b \end{align*}.

Simplify both of these equations and solve them simultaneously for a and b.
• February 3rd 2013, 12:48 AM
Oldspice1212
Re: Piece wise function help
Ah I think I missed a step, I got a= 7/2 and b = 13/2 this time hows that look?

9=10a-4b
2=8a-4b