Difficulty: Intermediate | Prerequisites: Partial derivatives (MATH 126 Midterm 1 material)
This material sits at the heart of multivariable optimisation. Once you can take partial derivatives, the next question is: where does a surface flatten out, and what happens there? Critical points and the Second Derivative Test answer that for functions of two variables. Level curves, meanwhile, give you a way to visualise a 3D surface on a 2D page, and reading them is a skill that shows up repeatedly in later topics (gradient fields, Lagrange multipliers, double integrals). If you are comfortable with first- and second-order partial derivatives, you are ready for this.
Critical points of f(x, y) occur where both partial derivatives equal zero (or are undefined). The Second Derivative Test uses the discriminant D = f_xx f_yy − (f_xy)² to classify each critical point as a local max, local min, or saddle point. Level curves are cross-sections of a surface at constant z-values, and their shape tells you the form of the function.
Critical point
A point (a, b) where both partial derivatives are zero: f_x(a, b) = 0 and f_y(a, b) = 0 (or where one or both partials are undefined). In simple terms, a critical point is where the surface levels out, with no upward or downward tilt in any direction.
Second Derivative Test (discriminant method)
A classification tool for critical points. Compute D = f_xx(a, b) · f_yy(a, b) − [f_xy(a, b)]². Think of D as a measure of how the surface curves around the critical point: it tells you whether the surface bends the same way in all directions (a max or min) or bends up in one direction and down in another (a saddle).
Saddle point
A critical point where the surface curves upward in one direction and downward in another, so it is neither a local max nor a local min. Think of it as the centre of a Pringles crisp: it dips down along one axis and rises along the other.
Level curve (contour line)
The set of all points (x, y) where f(x, y) = c for some constant c. In simple terms, a level curve is a horizontal slice through the surface at height c, projected down onto the xy-plane. If you have seen a topographic map, the contour lines are level curves of the elevation function.
Local maximum
A point where f is larger than at all nearby points. At the Second Derivative Test: D > 0 and f_xx < 0.
Local minimum
A point where f is smaller than at all nearby points. At the Second Derivative Test: D > 0 and f_xx > 0.
Compute the first-order partial derivatives f_x and f_y.
Set both equal to zero simultaneously: f_x = 0, f_y = 0. Solve the resulting system of equations for x and y. Each solution (a, b) is a critical point.
Compute the second-order partial derivatives: f_xx, f_yy, and f_xy.
Evaluate the discriminant at each critical point: D(a, b) = f_xx(a, b) · f_yy(a, b) − [f_xy(a, b)]².
Classify using the rules:
D > 0 and f_xx > 0: local minimum
D > 0 and f_xx < 0: local maximum
D < 0: saddle point
D = 0: the test is inconclusive (further analysis needed)
First partials:
f_x = 6y + 3x²
f_y = 6x − 2y
Setting f_y = 0 gives y = 3x. Substituting into f_x = 0:
6(3x) + 3x² = 0, so 18x + 3x² = 0, giving 3x(6 + x) = 0.
This yields x = 0 or x = −6.
x = 0 gives y = 0, so one critical point is (0, 0).
x = −6 gives y = −18, so the other critical point is (−6, −18).
Second partials:
f_xx = 6x
f_yy = −2
f_xy = 6
Discriminant: D = f_xx · f_yy − (f_xy)² = (6x)(−2) − 36 = −12x − 36.
At (0, 0): D = −36 < 0, so (0, 0) is a saddle point.
At (−6, −18): D = −12(−6) − 36 = 72 − 36 = 36 > 0, and f_xx(−6, −18) = 6(−6) = −36 < 0, so (−6, −18) is a local maximum.
A level curve f(x, y) = c is what you get when you slice a surface at height c and look straight down. Closely spaced level curves mean the surface is steep; widely spaced curves mean it is relatively flat.
The shape of the level curves tells you the form of the function:
Concentric circles centred at the origin: f depends only on x² + y² (e.g. f = x² + y²).
Straight lines: f is linear in x and y (e.g. f = 2x + 3y).
Parabolas opening in one direction: f involves a quadratic in one variable combined linearly with the other.
The exam showed level curves that are parabolas opening to the right, symmetric about the x-axis. These have the form x = ay² + c for various constants c, where a is fixed.
Setting f(x, y) = c and matching the shape x = ay² + c, you can rearrange to get c = x − ay². So a valid function is:
f(x, y) = x − y²
Other correct answers include f(x, y) = x − 2y², or any function of the form g(x − ay²) for fixed a > 0 and any monotone function g. The key insight is that x and y² must appear as a difference (x minus a constant times y²) so that the level curves solve to parabolas opening rightward.
D = f_{xx}(a,b) \cdot f_{yy}(a,b) - \bigl[f_{xy}(a,b)\bigr]^2\text{If } D > 0 \text{ and } f_{xx} > 0 \Rightarrow \text{local min}\text{If } D > 0 \text{ and } f_{xx} < 0 \Rightarrow \text{local max}\text{If } D < 0 \Rightarrow \text{saddle point}For level curves: set f(x, y) = c and solve for the curve equation in the xy-plane.
Critical-point analysis is how engineers find the most efficient operating conditions for a system, for instance the temperature and pressure that maximise yield in a chemical reactor. Level curves appear in weather maps (isobars are level curves of pressure) and topographic maps (contour lines are level curves of elevation).
Students often think that D > 0 alone is enough to declare a local min. It is not. You must also check the sign of f_xx to distinguish between a min (f_xx > 0) and a max (f_xx < 0).
Students sometimes forget to find all critical points. A system of two equations in two unknowns can have multiple solutions, and skipping a substitution step often drops one.
Confusing f_xy with f_xx or f_yy when computing D is a common algebraic slip. Write out each second partial separately before plugging into the formula.
For level curves, students sometimes set f(x, y) = 0 instead of f(x, y) = c. The constant c is arbitrary, and different values of c give different curves.
⚠️ The Second Derivative Test is nearly guaranteed on a MATH 126 midterm. Expect a function with two or three critical points requiring full classification.
⚠️ The system of equations step is where most marks are lost. Show every substitution and solution clearly.
⚠️ Level curve questions can appear as "give an example" or "match the function to its contour plot." Both require you to reason backward from a picture to a formula.
⚠️ If D = 0, say the test is inconclusive. Do not guess.
True or False: If D < 0 at a critical point, the point is a local minimum. (False, it is a saddle point.)
Fill in the blank: To find critical points, set f_x = ____ and f_y = ____. (Both equal to 0.)
True or False: f(x, y) = x + y² has parabolic level curves. (True, each level curve is y² = c − x, a parabola.)
True or False: D > 0 guarantees a local minimum. (False, you also need f_xx > 0.)
Fill in the blank: The discriminant formula is D = f_xx · f_yy − ____. ((f_xy)².)
Q: Find and classify the critical points of f(x, y) = x² + y² − 4x + 6y.
A: f_x = 2x − 4 = 0 gives x = 2. f_y = 2y + 6 = 0 gives y = −3. The only critical point is (2, −3). f_xx = 2, f_yy = 2, f_xy = 0. D = 4 > 0 and f_xx > 0, so (2, −3) is a local minimum.
Q: Give a function whose level curves are circles centred at the origin.
A: f(x, y) = x² + y². The level curves are x² + y² = c, which are circles of radius √c.
Q: Suppose D = 0 at a critical point. What can you conclude?
A: The Second Derivative Test is inconclusive. The point could be a local max, local min, or saddle point. You would need other methods (e.g. examining f directly near the point) to classify it.
Q: For f(x, y) = 6xy + x³ − y², explain why (0, 0) is a saddle point rather than a local extremum.
A: At (0, 0), D = −12(0) − 36 = −36 < 0. A negative discriminant means the surface curves upward in one direction and downward in another, ruling out a local max or min.
This connects to Lagrange multipliers, where you optimise a function subject to a constraint, because the unconstrained critical-point method here is the baseline before constraints are added. It also connects to gradient vectors: the gradient is zero at critical points, and level curves are always perpendicular to the gradient. Understanding level curves prepares you for double integrals, where you will describe regions in the xy-plane bounded by curves of this kind.
critical point, stationary point, second derivative test, discriminant, Hessian, saddle point, local maximum, local minimum, level curve, contour line, contour plot, multivariable optimisation, MATH 126, Calculus III, partial derivatives, classification of critical points, f_xx f_yy minus f_xy squared