<s>[INST] determine for which values of $x$ the function $f(x)=\begin{cases} \frac{x^2-4}{x-2}+10, &\text{if }x < 2\\ 2x^3-x, &\text{if } x ≥ 2\\ \end{cases}$ is continuous...
$$f(x)=\begin{cases}
\frac{x^2-4}{x-2}+10, &\text{if }x < 2\\
2x^3-x, &\text{if } x ≥ 2\\
\end{cases}$$
These kind of problems are very unclear to me, I only know these kind of functions have several graphs, in this example There are 2, But I can't figure out what these intervals means and even worse I don't even know where to start from.
 [/INST] As Zubin Mukerjee indicated in the comments, we must show that 


*

*$f$ is continuous when $x < 2$;

*$f$ is continuous when $x > 2$;

*$\lim_{x \to 2+} f(x) = \lim_{x \to 2-} f(x) = f(2)$.  


If $x < 2$, then
\begin{align*} 
f(x) & = \frac{x^2 - 4}{x - 2} + 10\\
     & = \frac{(x + 2)(x - 2)}{x - 2} + 10\\
     & = x + 2 + 10 && \text{provided $x \neq 2$}\\
     & = x + 12
\end{align*}
Since $x < 2$, we are not dividing by zero in the second step.  Hence, the given expression is equivalent to $x + 12$ in the interval $(-\infty, 2)$.  Since $x + 12$ is a polynomial, $f$ is continuous in the interval $(-\infty, 2)$.
If $x > 2$, then $f(x) = 2x^3 - x$.  Since this is a polynomial, $f$ is continuous the interval $(2, \infty)$.  
Since 
$$\lim_{x \to 2+} f(x) = \lim_{x \to 2+} (2x^3 - x) = 2 \cdot 2^3 - 2 = 16 - 2 = 14$$ 
and 
$$\lim_{x \to 2-} f(x) = \lim_{x \to 2-} (x + 12) = 2 + 12 = 14$$
and 
$$f(2) = 2 \cdot 2^3 - 2 = 16 - 12 = 14$$
it is true that 
$$\lim_{x \to 2+} f(x) = \lim_{x \to 2-} f(x) = f(2)$$
Since we have verified all three conditions, $f$ is continuous on $(-\infty, \infty)$.  
</s>