Re: Piece wise function help

This is extremely hard to read. Is your function $\displaystyle \displaystyle \begin{align*} f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} \textrm{ if } x < 2 \\ a x^2 - b x + 3 \textrm{ if } 2 \leq x < 3 \\ 4x - a + b \textrm{ if } x \geq 3 \end{cases} \end{align*}$?

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

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?

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.

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 $\displaystyle \displaystyle \begin{align*} a(2)^2 - b(2) + 3 = 2 \end{align*}$.

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 $\displaystyle \displaystyle \begin{align*} a(3)^2 - b(3) + 3 = 4(3) - a + b \end{align*}$.

Simplify both of these equations and solve them simultaneously for a and b.

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