Given $f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$When x = -3If f(x) is continuous for x = -3 then$\lim\limits_{x\rightarrow 3^{-}}f(x)=\lim\limits_{x\rightarrow -3^{+}}f(x)=f(-3)$Finding L.H.L.$\lim\limits_{x\rightarrow 3^{-}}|x|+3$$= \lim\limits_{h\rightarrow 0}|-3-h|+3$Putting h = 0 then we get,= | -3 - 0 | + 3= | -3 | + 3= 6Finding R.H.L.$\lim\limits_{x\rightarrow -3^{+}}-2x=\lim\limits_{h\rightarrow 0}-2(-3+h)$$= \lim\limits_{h\rightarrow 0}6-2h$Putting h = 0 then we get,= 6 - 2 $\times$ 0= 6Find f(x) at x = -3f(-3) = | -3 | + 3 = 6Hence, $\lim\limits_{x\rightarrow 3^{-}}f(x)=\lim\limits_{x\rightarrow -3^{+}}f(x)=f(-3)$Therefore, the function f(x) is continuous at x = -3.When x = 3If f(x) is continuous for x = 3 then$\lim\limits_{x\rightarrow 3^{-}}f(x)=\lim\limits_{x\rightarrow -3^{+}}f(x)=f(3)$Finding L.H.L.$\lim\limits_{x\rightarrow 3^{-}}-2x=\lim\limits_{h\rightarrow 0}-2(-3-h)$$= \lim\limits_{h\rightarrow 0}-6+2h$Putting h=0 then we get,= -6 + 2 $\times$ 0= -6Finding R.H.L.$\lim\limits_{x\rightarrow -3^{+}}6x+2=\lim\limits_{h\rightarrow 0}6(3+h)+2$$= \lim\limits_{h\rightarrow 0}18+6h+2$Putting h = 0 then we get,= 18 + 6 $\times$ 0 + 2= 20Find f(x) at x = 3f(3) = 6x + 2 at x = 3f(3) = 6 $\times$ 3 + 2 = 20Hence, $\lim\limits_{x\rightarrow 3^{-}}f(x) \ne \lim\limits_{x\rightarrow -3^{+}}f(x)=f(-3)$Therefore, the functionf(x) is discontinuous at x = 3When x < -3,For x < -3, f(x) = | x | + 3Since the function f(x) = | x | + 3 is a modulus function so it is continuous.∴ f(x) is continuous for x < -3When x > 3,For x > 3, f(x) = 6x + 2Since the function f(x) = 6x + 2 is a polynomial so it is continuous.∴ f(x) is continuous for x > 3When -3 < x < 3For -3 < x < 3, f(x) = -2xSince the function f(x) = -2x is a polynomial so it is continuous.∴ f(x) is continuous for -3 < x < 3∴ $f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$ discontinuous at x = 3 only.