Is $f(x)=\frac{1}{x}$ a real-valued function?
I think the definition of a real-valued function is that the range is in the real numbers. Is that right?
Hence, I think $f(x)=\frac{1}{x}$ is not a real-valued function.
If then, what are the other examples of real-valued functions?
Are polynomial functions the only real-valued functions?
A real valued is a function, whose range is $\mathbb{R}$ or some subset of $\mathbb{R}$.
$\frac{1}{x}$ for $x\in\mathbb{R}$ is real valued. We note that its range is $(-\infty,0)\cup (0,\infty)$, which is a subset of $\mathbb{R}$.
Polynomials are real valued, but they are not the only functions who are real valued.
A few examples of real valued functions:
The following are said to be real valued functions since their range is the set of real numbers, or some subset of the real numbers.
$f(x)=2$ ; The range is $\{2\}\subset \mathbb{R}$
$f(x)=11x$ ; The range is $\mathbb{R}$
$f(x)=e^x$ ; The range is $(0,\infty)\subset\mathbb{R}$
$f(x)=\dfrac{1}{x^2}$ ; The range is $(0,\infty)\subset \mathbb{R}$
etc...
The term "valued" refers to a function's codomain. real-valued, complex-valued, integer-valued, vector-valued, etc.
What is a real-valued function?
A real-valued function is a function that assigns real numbers to each member of its codomain or range.
Is $f(x)=\frac{1}{x}$ a real-valued function?
Yes because each member of the codomain of $f(x)=\frac{1}{x}$ is in fact a real number. Also note that excluding zero from the codomain does not change the fact that every other member is a real number. Symbolically, we have $$f:\mathbb R\setminus\{0\}\to\mathbb R\setminus\{0\}$$ Since $\mathbb R\setminus\{0\}$ is a subset of $\mathbb R$, then the expression above can be read as $f$ is a real-valued function of a real-valued variable.
A function which has either $\mathbb{R}$ or one of its subsets as its range is called a real valued function. Further, if its domain is also either $\mathbb{R}$ or a subset of $\mathbb{R}$ it is called a real function.
