Continuity on Interval

Question

I'm not understanding why $g(x)=x^2-3x+2$ is continuous over the interval (-4,4), and why $f(x)=\frac{1}{x}+3$ is NOT continuous over the interval (-7,7).

Any help would be appreciated. Thank you!

Answer 1

$$ f(x) \text{ continuous at $0$ } \iff \lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)=f(0)$$

But $f(0)$ doesn't exist, so this doesn't hold.

What's more, if it isn't continuous at $0$, it isn't continuous across a range which contains $0$, like $(-7,7)$ does.

Answer 2

$g(x)=x^2-3x+2$ is continuous everywhere and by extension over the interval $(-4, 4)$. It's continuous everywhere because it's a polynomial function. Polynomial functions are contentious over the entire real line.

$f(x)=\frac{1}{x}+3$ is not continuous over the interval $(-7,7)$ because $x=0$ (it lies right in the middle of the interval) is a point where this function is simply not defined. When $x=0$, you are dividing by zero which is not a legitimate operation in mathematics. Therefore, the function $f(x)=\frac{1}{x}+3$ can't have a value at that point. In other words, the value of the function $f(x)=\frac{1}{x}+3$ at the point $x=0$ does not exit. Graphically, this function looks discontinuous at that point (do you see how the graph of the function is not one single continuous line?):