# Continous Piecewise Function [closed]

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?

• Welcome to Math SE! Please modify your question to include your own thoughts/attempts at solving this problem. This site isn't a homework service and as it stands your question is unlikely to be meet with helpful responses. – DMcMor Oct 23 at 18:35

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.
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 $$d$$.