02 Multivariable Functions

A structured introduction to multivariable functions, including domains, graphs, level sets, limits, continuity, and a practical analysis workflow.

Functions, inputs, and outputs

A assigns exactly one output to each input from a specified set. For two variables, the standard notation is

f:D⊆R2→R,z=f(x,y).f:D\subseteq\mathbb{R}^2\to\mathbb{R},\qquad z=f(x,y).

Here, (x,y)(x,y) is the input, zz is the output, and DD is the . A function of three variables may be written w=f(x,y,z)w=f(x,y,z). The is the set of output values obtained from all inputs in the .

The variables may represent measurable quantities such as position, time, temperature, pressure, concentration, or elevation. For example, T(x,y)T(x,y) can represent temperature over a flat plate, while p(x,y,z)p(x,y,z) can represent pressure at a point in space.

To evaluate a function, substitute the input values into its formula. If

f(x,y)=x2+3xy−y2,f(x,y)=x^2+3xy-y^2,

then

f(1,2)=12+3(1)(2)−22=3.f(1,2)=1^2+3(1)(2)-2^2=3.

A formula can impose restrictions on its inputs. For example, f(x,y)=9−x2−y2f(x,y)=\sqrt{9-x^2-y^2} is real-valued only when x2+y2≤9x^2+y^2\le 9, so its is the closed disk of radius 33 centered at the origin.

Takeaway: Identify the inputs, output, , and before evaluating or interpreting a multivariable function.

Finding domains

The consists of every input point for which the formula is defined. For a function of two variables, the is a region in the xyxy-plane. Check restrictions systematically:

  • A denominator must be nonzero. For f(x,y)=1x−yf(x,y)=\frac{1}{x-y}, require x−y≠0x-y\ne 0. The is the plane excluding the line y=xy=x.

  • The radicand of an even root must be nonnegative. For g(x,y)=4−x2−y2g(x,y)=\sqrt{4-x^2-y^2}, require x2+y2≤4x^2+y^2\le 4, which is the disk of radius 22.

  • The argument of a logarithm must be positive. For h(x,y)=ln⁡(x+y)h(x,y)=\ln(x+y), require x+y>0x+y>0, which describes a half-plane.

When several restrictions occur, impose all of them at once. For

k(x,y)=x−yx+y,k(x,y)=\frac{\sqrt{x-y}}{x+y},

the conditions are

x−y≥0andx+y≠0.x-y\ge 0\qquad\text{and}\qquad x+y\ne 0.

Thus, the is the region on or below the line y=xy=x, with points on the line y=−xy=-x removed.

Takeaway: Translate algebraic restrictions into geometric regions, and use the intersection of all restrictions as the .

Graphs and vertical traces

The graph of a function of two variables is the set

{(x,y,z)∈R3:z=f(x,y)}.\{(x,y,z)\in\mathbb{R}^3:z=f(x,y)\}.

It is generally a surface rather than a plane curve. For

f(x,y)=x2+y2,f(x,y)=x^2+y^2,

the graph is

z=x2+y2.z=x^2+y^2.

Because z≥0z\ge 0, the surface has a lowest point at (0,0,0)(0,0,0) and rises as the distance from the origin increases. It is an upward-opening circular paraboloid.

A is found by fixing one input variable. Setting x=0x=0 gives

z=y2,z=y^2,

which is a parabola in the yzyz-plane. Setting y=0y=0 gives

z=x2,z=x^2,

which is a parabola in the xzxz-plane. Other fixed values, such as x=ax=a, produce additional cross-sections that help reveal the surface's shape.

Takeaway: Use the graph equation to understand the full surface, and use vertical traces to reduce its analysis to familiar two-dimensional curves.

Level curves and level surfaces

A for a constant cc is found by setting

f(x,y)=c.f(x,y)=c.

It lies in the xyxy-plane and connects points where the function has the same value. For the paraboloid f(x,y)=x2+y2f(x,y)=x^2+y^2, the level curves satisfy

x2+y2=c.x^2+y^2=c.

For c>0c>0, these are circles centered at the origin with radius c\sqrt{c}. In particular, c=1c=1 gives a circle of radius 11, c=4c=4 gives a circle of radius 22, and c=0c=0 gives only the point (0,0)(0,0). A collection of level curves forms a . Closely spaced curves indicate relatively rapid change, while widely spaced curves indicate more gradual change.

For the plane f(x,y)=2x−yf(x,y)=2x-y, the level curves satisfy

2x−y=c,2x-y=c,

or equivalently,

y=2x−c.y=2x-c.

They are parallel lines with slope 22.

For a function of three variables, a is found by setting

f(x,y,z)=c.f(x,y,z)=c.

For f(x,y,z)=x2+y2+z2f(x,y,z)=x^2+y^2+z^2, the level surfaces are

x2+y2+z2=c,x^2+y^2+z^2=c,

which are spheres centered at the origin with radius c\sqrt{c} when c>0c>0.

A of f(x,y)f(x,y) can also be viewed as the intersection of the graph z=f(x,y)z=f(x,y) with the horizontal plane z=cz=c.

Takeaway: Set the function equal to a constant to identify equal-value sets: curves in two input dimensions and surfaces in three input dimensions.

Limits and path testing

The limit

lim⁡(x,y)→(a,b)f(x,y)=L\lim_{(x,y)\to(a,b)}f(x,y)=L

means that function values approach LL as the point (x,y)(x,y) approaches (a,b)(a,b) from within the . The point can be approached along infinitely many paths. The distance to the target point is

(x−a)2+(y−b)2,\sqrt{(x-a)^2+(y-b)^2},

so approaching (a,b)(a,b) means that this distance approaches zero.

Direct substitution often works for expressions assembled from continuous operations when the input is in the valid . For example,

lim⁡(x,y)→(1,2)(x2+3xy+y2)=12+3(1)(2)+22=11.\lim_{(x,y)\to(1,2)}(x^2+3xy+y^2)=1^2+3(1)(2)+2^2=11.

To test whether a limit exists, compare paths when the expression is not plainly continuous. Consider

f(x,y)=x2−y2x2+y2.f(x,y)=\frac{x^2-y^2}{x^2+y^2}.

Along y=0y=0,

f(x,0)=x2x2=1,f(x,0)=\frac{x^2}{x^2}=1,

whereas along x=0x=0,

f(0,y)=−y2y2=−1.f(0,y)=\frac{-y^2}{y^2}=-1.

Since the paths give different values, the limit at (0,0)(0,0) does not exist. This is . Two paths with different results are enough to disprove a limit, but agreement along a few paths is not enough to prove one.

The formal definition states that for every ε>0\varepsilon>0, there is a δ>0\delta>0 such that

0<(x−a)2+(y−b)2<δ0<\sqrt{(x-a)^2+(y-b)^2}<\delta

implies

∣f(x,y)−L∣<ε.\lvert f(x,y)-L\rvert<\varepsilon.

In words, inputs sufficiently close to (a,b)(a,b) must produce outputs as close to LL as desired.

Takeaway: A multivariable limit must be independent of the path of approach. Use direct substitution when continuity is clear and path comparisons when it is not.

Continuity and removable discontinuities

A function is a point when its value, nearby limiting behavior, and limiting value agree. At (a,b)(a,b), verify all three conditions:

  1. f(a,b)f(a,b) is defined.

  2. lim⁡(x,y)→(a,b)f(x,y)\lim_{(x,y)\to(a,b)}f(x,y) exists.

  3. lim⁡(x,y)→(a,b)f(x,y)=f(a,b)\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b).

Polynomials are continuous everywhere. Sums, differences, products, compositions, valid roots, logarithms on their positive-argument domains, trigonometric functions, and rational functions with nonzero denominators are continuous wherever they are defined.

For example,

f(x,y)=cos⁡(x2+3y2)f(x,y)=\cos(x^2+3y^2)

is continuous everywhere because the polynomial inside the cosine is continuous and cosine is continuous for every real input. The function

g(x,y)=ln⁡(x+y)x2−yg(x,y)=\frac{\ln(x+y)}{x^2-y}

is continuous on the region

x+y>0,qquadx2−y≠0.x+y>0,\\qquad x^2-y\ne 0.

A removable discontinuity occurs when a limit exists but does not equal the function's assigned value. Consider

f(x,y)={x2−y2x−y,x≠y,0,x=y.f(x,y)= \begin{cases} \dfrac{x^2-y^2}{x-y}, & x\ne y,\\[6pt] 0, & x=y. \end{cases}

For x≠yx\ne y, factor and simplify:

x2−y2x−y=x+y.\frac{x^2-y^2}{x-y}=x+y.

Therefore,

lim⁡(x,y)→(1,1)f(x,y)=lim⁡(x,y)→(1,1)(x+y)=2,\lim_{(x,y)\to(1,1)}f(x,y)=\lim_{(x,y)\to(1,1)}(x+y)=2,

but f(1,1)=0f(1,1)=0. The function is not (1,1)(1,1). Assigning the value 22 at that point would remove the discontinuity.

Takeaway: Continuity combines , limits, and function values. Always check that the point is in the before declaring continuity.

A practical analysis workflow

A reliable analysis follows a fixed sequence:

  1. Identify the variables and output. Decide whether the function maps from R2\mathbb{R}^2, R3\mathbb{R}^3, or another space into R\mathbb{R}.

  2. Find the . Check denominators, even roots, logarithms, and other restrictions.

  3. Describe the graph. Use z=f(x,y)z=f(x,y), recognizable surfaces, and vertical traces.

  4. Find level curves or level surfaces by setting the function equal to a constant cc.

  5. Evaluate limits. Try direct substitution when the expression is continuous-looking; otherwise simplify, compare paths, or use bounds or polar coordinates.

  6. Check continuity. Confirm that the function is defined, that the limit exists, and that the limit equals the function value.

This workflow connects algebraic formulas with geometric meaning. restrictions determine where the function exists, graphs show its overall shape, level sets display equal values, limits describe local behavior, and continuity determines whether the local behavior matches the function's assigned value.

Final takeaway: For any multivariable function, move from inputs and to geometry, then to local behavior through limits and continuity.