Given a function $f$ defined by $$f(x) := \left( x^\sigma + b \right)^{1/\sigma}, \qquad (\sigma <0, \quad b \in \mathbb{R}_+).$$
Since $\sigma <0$ and thus the negative exponent (power) functions $x^\sigma$ (or $x^{1/\sigma}$) are not well defined for $x=0$, it seems like this function $f$ is also not well defined for $x=0$.
However, when I draw the graph of this function $f$, I found the graph of $f$ could reach $(0,0)$ (i.e., $f(0)=0$), so that I am confused whether such a function $f$ is well defined on zero point.
Question 1: It is clear that the above function $f$ is well defined on $(0, \infty)$, but I am wondering that is $f$ well defined on $x =0$ point?
Question 2: If $f$ is well defined at $x=0$ (and hence well defined on $\mathbb{R}_+$), then is this function $f$ continuous at $x=0$? If so, how to prove it?
Any idea or suggestions are most welcome and much appreciated!
Thank you in advance!
$f(0)$ is not defined because you cannot calculate $0^\sigma$ as you say. However you can calculate $\lim_{x\to 0^+}f(x)$. Intuitively, as $x$ gets very small $x^\sigma$ will get very large so $b$ will not matter. You will then have $f(x)$ very close to $x$ for small $x$ and the limit will be $0$. If you want, you could define a new function $g(x)$ by $$g(x)=\begin {cases} 0&x=0\\f(x)&x \gt 0 \end {cases}$$ Now $g(x)$ is well defined and continuous from above at $0$. It is the same idea as a removable singularity in a function, where you "fill in a hole" of the definition in such a way that you make the function continuous.