Okay this one is different. when I take the derivative I assume I can just plug in but I am not sure if it will work.

f(x)=

$\displaystyle ax^2 + 1/3$, x>1

$\displaystyle bx-10/3$, x <1

I took the limit, and got $\displaystyle a +1/3 = b - 10/3$, then I took the derivative of the equation and took the same limit and got 2a = b. Do I just plug 2a in for b?