Side continuity

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.