Verify if the following functions are continuous:
a) $$\displaystyle f(x)=\left\{\begin{array}{rcl} 2x+1 & \mbox{ if } & x<1 \\ 3x & \mbox{ if } & x \geq 1\end{array}\right.$$
b) $$\displaystyle f(x)=\left\{\begin{array}{rcl} \frac{1}{x} & \mbox{ if } & x\neq 0 \\ 0 & \mbox{ if } & x =0 \end{array}\right.$$
c) $$\displaystyle f(x)=\left\{\begin{array}{rcl} x^2+1 & \mbox{ if } & -1 < x < 1 \\ 3x & \mbox{ if } & x \geq 1 \mbox{ or } x \leq -1 \end{array}\right.$$
Development:
a) The functions that define $$f(x)$$ are continuous, so the only point where we could have some problem is at $$x=1$$, where the two subfunctions meet: $$$\displaystyle \begin{array} {l} \lim_{x \to 1^+}f(x)=\lim_{x \to 1} 3x= 3 \\ \lim_{x \to 1^-}f(x)=\lim_{x \to 1} (2x+1)= 2+1= 3 \\ f(1)=3 \end{array}$$$ and since the side limits coincide with the value of the function, the function is continuous.
b) The function $$\dfrac{1}{x}$$ is continuous in its domain. We need to verify if $$f(x)$$ is continuous at zero: $$$\displaystyle \begin{array} {l} \lim_{x \to 0^+}f(x)=\lim_{x \to 0^+} \dfrac{1}{x}= +\infty \\ \lim_{x \to 0^-}f(x)=\lim_{x \to 0^-} \dfrac{1}{x}= -\infty \\ f(0)=0 \end{array}$$$ Therefore the limits do not coincide with the function at the zero; the function is not continuous.
c) The functions that define $$f(x)$$ are continuous so we only need to verify the points $$x=1$$ and $$x =-1$$, where the different subfunctions meet:
Continuity at $$x=1$$: $$$\displaystyle \begin{array} {l} \lim_{x \to 1^+}f(x)=\lim_{x \to 1} 2x= 2 \\ \lim_{x \to 1^-}f(x)=\lim_{x \to 1} (x^2+1)= 1+1= 2 \\ f(1)=2 \end{array}$$$ therefore the function is continuous at $$x=1$$.
Continuity at $$x =-1$$: $$$\displaystyle \begin{array} {l} \lim_{x \to -1^+}f(x)=\lim_{x \to -1} (x^2+1)= (-1)^2+1=2 \\ \lim_{x \to -1^-}f(x)=\lim_{x \to -1} 2x= -2 \\ f(-1)=-2 \end{array}$$$ therefore the function is not continuous at $$x =-1$$, so we will not have a continuous function.
Solution:
a) Continuous function
b) Discontinuous function
c) Discontinuous function