If a function $f$ is continuous over its domain , so can we say that $f^k$ will also be continuous over the same domain, where k is any real number? I'm working out a problem of proving a function continuous by the help of continuous functions,there I stuck at the last step which requires $f^k$ to be continuous if $f$ is continuous over its domain
 A: No, consider $f:\mathbb{R}\to \mathbb{R}$ given by  $$f(x)=x$$ which is continuous, then consider $k=-1$ then $$f^{k}(x)=\frac{1}{x}$$ which is not continuous in $\mathbb{R}$
A: Your main problem is to be sure that $f(x)^k$ is defined everywhere. Generally the function $x\mapsto x^k$ is continuous in its domain, so if you can compose it with $f$, then the result will be continuous too.
If $f(x)>0$ everywhere, then you're good no matter what $k\in\mathbb R$ is.
Otherwise you'll need to consider several cases for $k$:

*

*$k$ is an integer $\ge 0$ or positive rational with odd denominator: Always OK no matter what the range of $f$ is.

*$k$ is any other positive real (i.e. irrational or with even denominator): You need $f(x)\ge 0$ everywhere.

*$k$ is a negative integer or rational with odd denominator: You need $f(x)\ne 0$ everywhere. If the domain of $f$ is connected, this implies that it is either all positive or all negative.

*$k$ is any other negative real: You need $f(x) > 0$ everywhere.

