# Is $f(x) = x^2$ a scalar function?

Take a simple parabola. It is a function that has a one-dimensional co-domain $$f(x) = x^2$$

It is mapping the set of values in its domain, to one-dimensional values in its co-domain, and it satisfies the definition of a scalar function.$$f : \mathbb{R^n} \rightarrow \mathbb{R}$$

Is it a scalar function? I ask this as generally when we are initially taught simple functions such as this, it is never put in the context of scalar functions.

### Sub-question

But we can also set $$y= f(x)$$ and plot the function on a Cartesian Plane (two-dimensional Euclidean Space) But it "seems" that by doing this (plotting the function) we are relating a set of $x$-values on the number line to a set of $(x, f(x))$ values in a two-dimensional Euclidean Space.

But the function itself is not mapping one-dimensional Euclidean Space to two-dimensional Euclidean Space, as it just producing scalar outputs.

Am I correct in saying that the act of plotting the function with its corresponding $x$-values is what is mapping one-dimensional Euclidean Space onto two-dimensional Euclidean Space, as the function itself is definitely not doing that. Is the plotting what is mapping one-dimensional Euclidean Space onto two-dimensional Euclidean Space? $$plot: \mathbb{R} \rightarrow \mathbb{R^2}$$

Any function $f:\mathbb R^n\to\mathbb R$ is termed a scalar-valued function, even if $n=1$. So $f(x)=x^2$ is a scalar-valued function. You are also correct in differentiating a function $\mathbb R\to\mathbb R$ from its graph in $\mathbb R^2$. The function $f:\mathbb R\to\mathbb R$ is scalar valued, but we can use it to define a function $F:\mathbb R\to\mathbb R^2$ by $F(x)=(x,f(x))$, which is not scalar valued. The function $F$ is vector-valued, and gives the graph, or "plot" of $f$.
You are correct in saying that the act of plotting the function with its corresponding $x$-values is mapping one-dimensional Euclidean space into two-dimensional Euclidean space, and this mapping is formalized by the function $F$ described above.