Continous Piecewise Function I recently found this question: find $c$ and $d$ such that
$$f(x) = \begin{cases} 
      cx+4d & x<2\\
      x^{2}+4 & 2\leq x\leq 3 \\
      dx^{2}+\frac{2x}{c}+1 & x>3
   \end{cases}$$
is continuous everywhere.
How should I solve this?
 A: Note that each piece of the function is continuous, so you only need to check if the different pieces are continuous at each junction. You can do this by checking if the limits approached from the right and left are equal.
If you are still stuck, here is the solution:

 Since all "pieces" of the function are continuous, it remains that the junctions between pieces must be continuous. We must have:

 $$\lim_{x\rightarrow 2^{-}}(cx + 4d) = \lim_{x\rightarrow 2^{+}}( x^{2} + 4)$$
$$\lim_{x\rightarrow 3^{-}}(x^{2} + 4) = \lim_{x\rightarrow 3^{+}}(dx^{2} + \frac{2x}{c} + 1)$$

 Evaluating the limits:

 $$2c + 4d = 8$$
$$13 = 9d + \frac{6}{c} + 1$$

 Thus:

 $$c = 4-2d$$
$$13 = 9d + \frac{3}{2-d}+1$$
$$12(2-d) = 9d(2-d) + 3$$
$$9d^{2} - 30d + 21 = 0$$
$$3(3d-7)(d-1) = 0$$
$$d = 1,\frac{7}{3}$$

 Then, substituting:

 $$\boxed{(c,d) = (2,1),(-\frac{2}{3},\frac{7}{3})}$$

 We may verify that both pairs solve the original system of equations.



A: Use middle definition to get values $f(2)=8$ and   $f(3)=13$.  Then get a pair of equations from continuity:  $2c+4d=8$ and $9d+\frac{6}{c}+1=13$.  Solve for $c$ and $d$.
