Continuity of composite function

Question

$f:\mathbb{R}\to\mathbb{R}$ is a continuous function.

I need to check whether $g(x)$ is also continuous for: $$ g(x)=\frac{1}{\min\{f(x),-1\}} $$ Two questions:

1. Can I show the continuity of $g(x)$ by using $f(x)=x$ or should I use more generalized approach?
2. Am I right that $g(x)$ will be continuous, as composition of two continuous functions ($min\{\}$ and $f(x)$) is also continuous and $\frac{1}{x}$ is continuous for $(-\infty,-1)$ ?

Answer 1

You'll need to be more general. If you know that $\min\{f(x),h(x)\}$ is continuous whenever $f(x),h(x)$ are, then $\min\{f(x),-1\}$ is a continuous non-zero function, and so $g$ is the reciprocal of a non-zero continuous function, so continuous.

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As a hint for how to show that $\min\{f(x),h(x)\}$ is continuous if $f(x),h(x)$ are, note that we have $$\min\{a,b\}:=\frac{a+b-|a-b|}2$$ for any real $a,b.$ (Why?) Thus, $$\min\{f(x),h(x)\}=\frac{f(x)+h(x)-|f(x)-h(x)|}2.$$ Some basic continuity results should let you prove the desired result.

Answer 2

1. No. When proving continuity, the function $f$ is an arbitrary, continuous function. You could have chosen $f$ to be whatever you like, were you trying to generate a counter example.
2. Although the intuition is right, your statement is weird. For example, how exactly would you define the function $\min$? And how would you prove it's continuous? Also, it doesn't help you much saying that $\frac{1}{x}$ is continuous on $(-\infty, -1)$, since you have to prove continuity $\forall x\in \mathbb{R}$.

Finally, as @Cameron Buie suggested, you might want to try a more general approach. Start with proving (by definition, even) that $\min{\{f(x),-1\}}$ is continuous. Then show that $\min{\{f(x),-1\}}\neq 0$ $\forall x\in \mathbb{R}$. And then conclude the continuity of $g(x)$.