# Graph of a function of multiple variables

## Definition

Suppose $f$ is a real-valued function of $n$ variables $x_1,x_2,\dots,x_n$. The graph of $f$ is a subset of $\R^{n+1}$, with coordinate axes $x_1,x_2,\dots,x_n,x_{n+1}$, given as follows:

$\{ (x_1,x_2,\dots,x_n,f(x_1,x_2,\dots,x_n)) \mid (x_1,x_2,\dots,x_n) \in \operatorname{dom}(f)\}$

where $\operatorname{dom}(f)$ denotes the domain of $f$.

Alternatively, it is given by the equation:

$x_{n+1} = f(x_1,x_2,\dots,x_n)$

For nice enough functions, this graph looks like a hypersurface of codimension one (and dimension $n$) inside $\R^{n+1}$.

## Aspects

### Domain and range

Aspect of the function How it can be deduced from the graph
domain project the entire graph on the $x_1x_2 \dots x_n$-hyperplane
range project the entire graph on the $x_{n+1}$-axis

### Vertical line test

The vertical line test for a function of one variable says that every vertical line intersects the graph in exactly one point if the $x$-coordinate is in the domain and in no point if the $x$-coordinate is not in the domain. There is an analogous test for a function of multiple variables. This says that for any line parallel to the $x_{n+1}$-axis, the intersection with the graph has size one (if the intersection with the $x_1x_2\dots x_n$-hyperplane is in the domain) or zero (if it isn't).

## Restriction to one variable

### Graph of the restriction

We can restrict the function to a function of fewer variables by fixing the value of some of the variables. Suppose $A$ is a subset of \{ 1,2,\dots,n\}[/itex]. Suppose we fix the values of the coordinates in $A$, so we get $x_j = a_j$ for $j \in A$.

This is now a function of the remaining variables, which is a total of $n - |A|$ variables.

The graph of this function is obtained by intersecting the original graph with the affine subspace given by $x_j = a_j, j \in A$. Note that this subspace has dimension $n + 1 - |A|$, and the intersection of the graph with this is expected to have dimension $n - |A|$.

Here's an extreme case: $A$ has size $n - 1$, and the only variable omitted is $i$. Then, the function we obtain is a function of one variable:

$x_i \mapsto f(a_1,a_2,\dot,a_{i-1},x_i,a_{i+1},\dots,a_n)$

The graph of this is obtained by intersecting the original graph with the plane given by equations $x_j = a_j$ for all $j \ne i$. There are $n - 1$ such equations. The plane itself is parallel to the $x_ix_{n+1}$-plane.

### Continuity in each variable and separate continuity in graphical terms

We have the following:

Assertion about continuity How we can verify it from the graph
$f$ is continuous in $x_i$ at the point $(a_1,a_2,\dots,a_n)$ Consider the graph restricted to the plane $x_j = a_j, j \ne i$. This graph is continuous at $x_i = a_i$.
$f$ is separately continuous in all variables at the point $(a_1,a_2,\dots,a_n)$. The above holds for all $i \in \{ 1,2,3,\dots,n\}$.
$f$ is continuous in $x_i$ everywhere. The restrictions of the graph to all planes parallel to the $x_ix_{n+1}$-plane are continuous functions.
$f$ is separately continuous in all variables everywhere. The above holds for all $i \in \{ 1,2,3,\dots,n\}$.

### Partial derivatives in graphical terms

For further information, refer: partial derivative

Suppose $f$ is a function of $n$ variables $x_1,x_2,\dots,x_n$ and suppose $(a_1,a_2,\dots,a_n)$ is a point in the domain of $f$. Consider the graph of $f$ in $\R^{n+1}$ given by:

$x_{n+1} = f(x_1,x_2,\dots,x_n)$

For any $i \in \{ 1,2,\dots,n\}$, we define the partial derivative $f_{x_i}(a_1,a_2,\dots,a_n)$, also denoted $f_i(a_1,a_2,\dots,a_n)$, as follows:

• First, consider the intersection of the graph of $f$ with the plane given by the set of $n - 1$ equations $x_j = a_j$ for all $j \ne i$. This is a plane parallel to the $x_ix_{n+1}$-plane.
• In this plane, consider the slope of the tangent line at $x_i =a_i$. This is the value of the partial derivative.

### Gradient vector in graphical terms

For further information, refer: gradient vector

Suppose $f$ is a function of multiple variables $x_1,x_2,\dots,x_n$ and suppose $(a_1,a_2,\dots,a_n)$ is a point in the domain of $f$. We say that $f$ is differentiable at $(a_1,a_2,\dots,a_n)$ if the gradient vector $(\nabla f)(a_1,a_2,\dots,a_n)$ exists. This is equivalent to the graph of the function having a well defined tangent hyperplane at the point $(a_1,a_2,\dots,a_n,f(a_1,a_2,\dot,a_n))$. The equation of the tangent hyperplane is given by:

$x_{n+1} - f(a_1,a_2,\dots,a_n) = (\nabla f)(a_1,a_2,\dots,a_n) \cdot (\langle x_1,x_2,\dots,x_n\rangle - \langle a_1,a_2,\dots,a_n \rangle)$