Critical Points, Tangent Planes & Local Extrema Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice multivariable local shape: finding critical points from \(\nabla f=0\), writing tangent planes, linearizations, and normal vectors, applying the two-variable Hessian determinant \(D=f_{xx}f_{yy}-f_{xy}^2\), classifying positive definite, negative definite, and indefinite Hessians, handling inconclusive \(D=0\) cases, checking boundary and compact-set extrema, and using Lagrange multipliers for regular constraints. Open the lesson for short worked examples and quick checks.
Answer the question set and review your mistakes at the end.
How this local extrema practice works
1. Take the practice set: answer questions about gradients, tangent planes, Hessians, constrained extrema, and compactness.
2. Open the lesson: review the definitions, recognition tests, worked examples, and single-answer checks.
3. Retry: return to the question set and first decide whether the problem is asking for a point, a plane, a classification, or a global comparison.
What you will learn in the critical points, tangent planes, and local extrema lesson
Critical points and first-order tests
Interior differentiable extrema: \(\nabla f(a)=0\) is necessary
Critical point: gradient zero or derivative information unavailable in the domain
Solve \(f_x=0\) and \(f_y=0\), then classify instead of assuming an extremum
Critical Points, Tangent Planes & Local Extrema Lesson
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Lesson overview
Purpose: Build a reliable workflow for local shape in two or more variables: find critical points, write tangent planes, classify local minima, local maxima, and saddle points, handle flat or degenerate cases, and compare candidates on constrained or compact domains.
Success criteria
State the first-order necessary condition for an interior differentiable local extremum.
Find critical points by solving \(f_x=0\) and \(f_y=0\).
Write the tangent plane to \(z=f(x,y)\) at \((a,b)\).
Find normal vectors for graph and level-surface tangent planes.
Use \(D=f_{xx}f_{yy}-f_{xy}^2\) to classify ordinary two-variable critical points.
Read Hessian definiteness from signs or eigenvalues.
Recognize why \(D=0\) is inconclusive and test signs along curves when needed.
Use compactness and boundary checks for global extrema.
Apply \(\nabla f=\lambda\nabla g\) only at regular constrained candidates.
Key vocabulary
Critical point: a point where \(\nabla f=0\) or where derivative information needed for the test is not available.
Tangent plane: the affine first-order model of a differentiable graph \(z=f(x,y)\).
Hessian: the matrix of second partial derivatives.
Positive definite: quadratic form is positive in every nonzero direction.
Saddle point: nearby values occur both above and below the value at the point.
Regular constraint: a constraint \(g=c\) with \(\nabla g≠ 0\) at the candidate.
Quick pre-check
Pre-check: At an interior local extremum of a differentiable function \(f(x,y)\), what must hold?
Hint: A differentiable interior extremum has zero first-order change in every direction.
Critical points are candidates, not automatic extrema
Learning goal: Find the candidate points where local extrema or saddles can occur, then keep classification separate from discovery.
Key idea
For a differentiable scalar function \(f(x,y)\), an interior local maximum or minimum must satisfy \(\nabla f(a,b)=(0,0)\). This condition is necessary, not sufficient: \(x^2-y^2\) and \(xy\) both have critical points at the origin but are saddles.
Recognition checklist
Check that the point is inside the domain, not just on a boundary.
Compute \(f_x\) and \(f_y\).
Solve the system \(f_x=0,\ f_y=0\).
Add any points where the derivatives fail but the function is still in the domain.
Classify each candidate using Hessians, signs, or boundary comparisons.
Worked example
Example: Find the critical point of \(f(x,y)=x^2+y^2-2x+4y\).
The gradient is \(\nabla f=(2x-2,2y+4)\). Setting both components to zero gives \(x=1\) and \(y=-2\), so the only critical point is \((1,-2)\). Its Hessian is \(2I\), so it is a strict local minimum.
Try it
Try it: For \(f(x,y)=x^2+y^2-4y\), where is the critical point?
Hint: \(\nabla f=(2x,2y-4)\).
Tangent planes are first-order models
Learning goal: Use first partial derivatives to build the local affine approximation to a graph or level surface.
Key idea
If \(f\) is differentiable at \((a,b)\), then the tangent plane to \(z=f(x,y)\) is \[z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b).\] Equivalently, \(f(a+h,b+k)\approx f(a,b)+f_x(a,b)h+f_y(a,b)k\).
Tangent-plane patterns
For a graph \(z=f(x,y)\), compute the value and two first partials at the base point.
A normal vector to that graph is \((f_x(a,b),f_y(a,b),-1)\); the opposite vector is also normal.
If \(\nabla f(a,b)=0\), the graph tangent plane is horizontal: \(z=f(a,b)\).
For an implicit surface \(F(x,y,z)=0\), the normal vector is \(\nabla F(a,b,c)\) when it is nonzero.
A tangent plane is a local approximation, not a global equality unless the function is already affine.
Worked example
Example: Write the tangent plane to \(z=x^2+y\) at \((1,2)\).
Here \(f(1,2)=3\), \(f_x=2x\), so \(f_x(1,2)=2\), and \(f_y=1\). Thus \[z=3+2(x-1)+(y-2).\] The matching linear approximation is \(f(1+h,2+k)\approx3+2h+k\).
Try it
Try it: For \(z=x^2+y^2\), what is the tangent plane at \((0,0)\)?
Hint: At the origin, the value and both first partials are \(0\).
The Hessian tells the local quadratic shape
Learning goal: Use \(D=f_{xx}f_{yy}-f_{xy}^2\) to classify nondegenerate two-variable critical points.
Key idea
At a critical point of a \(C^2\) function \(f(x,y)\), the second-order Taylor part is controlled by the Hessian. In two variables, the determinant \(D=f_{xx}f_{yy}-f_{xy}^2\) separates the usual positive definite, negative definite, indefinite, and inconclusive cases.
Two-variable test rules
If \(D>0\) and \(f_{xx}>0\), the point is a strict local minimum.
If \(D>0\) and \(f_{xx}<0\), the point is a strict local maximum.
If \(D<0\), the Hessian is indefinite and the point is a saddle.
If \(D=0\), the test is inconclusive.
Worked example
Example: Classify the origin for \(f(x,y)=x^2+xy+y^2\).
The origin is critical. Here \(f_{xx}=2\), \(f_{yy}=2\), and \(f_{xy}=1\), so \(D=2\cdot2-1^2=3>0\). Since \(f_{xx}>0\), the origin is a strict local minimum.
Try it
Try it: If \(D=f_{xx}f_{yy}-f_{xy}^2<0\) at a critical point, what does the test give?
Hint: Negative determinant means the Hessian curves up in some directions and down in others.
Eigenvalue signs give the same local shape in any dimension
Learning goal: Read local behavior from Hessian definiteness, especially beyond two variables.
Key idea
At a critical point, a positive definite Hessian gives a strict local minimum, a negative definite Hessian gives a strict local maximum, and an indefinite Hessian gives a saddle. For symmetric Hessians, this can be read from eigenvalue signs: all positive, all negative, or mixed signs.
Sign patterns
All Hessian eigenvalues positive: positive definite.
All Hessian eigenvalues negative: negative definite.
Both positive and negative Hessian eigenvalues: indefinite.
Zero eigenvalues: the quadratic test may be inconclusive and higher-order terms can decide.
Worked example
Example: Suppose a critical point has Hessian eigenvalues \(-2\) and \(5\). What type of point is suggested?
The signs are mixed, so the Hessian is indefinite. The quadratic approximation has both positive and negative directions, so the critical point is a saddle.
Try it
Try it: If the Hessian eigenvalues at a critical point are \(2\) and \(5\), what does the second-order test indicate?
Hint: Both eigenvalues are positive.
When the Hessian test is silent, test the function directly
Learning goal: Handle \(D=0\) or zero Hessian cases without overclaiming.
Key idea
The statement \(D=0\) means the standard two-variable second derivative test gives no classification. It does not mean minimum, maximum, or saddle. Look at signs of \(f(x,y)-f(a,b)\) along simple paths, or use algebraic factorization and higher-order terms.
What to do when the test fails
Check coordinate axes and simple lines such as \(y=x\) and \(y=-x\).
Look for expressions that are always nonnegative or always nonpositive.
Remember that a local minimum can be non-strict, as in \(f(x,y)=x^2\).
If different paths give values on both sides of \(f(a,b)\), the point is a saddle.
Worked example
Example: Classify the origin for \(f(x,y)=x^4-y^4\).
The Hessian at the origin is the zero matrix, so the quadratic test is inconclusive. Along the \(x\)-axis, \(f(x,0)=x^4>0\) for \(x≠0\). Along the \(y\)-axis, \(f(0,y)=-y^4<0\) for \(y≠0\). Values occur on both sides of \(0\), so the origin is a saddle.
Try it
Try it: For \(f(x,y)=x^2\), what is the origin?
Hint: \(x^2\ge0\), but the function equals \(0\) along the whole \(y\)-axis.
Global extrema require candidates from every part of the domain
For \(f\) constrained by \(g(x,y)=c\), a regular constrained extremum satisfies \(\nabla f=\lambda\nabla g\). The word regular matters: \(\nabla g≠0\) at the candidate. For global extrema on a compact set, compare all interior candidates, boundary candidates, and any corners or singular points.
Global-extrema workflow
Use \(\nabla f=0\) for interior differentiable candidates.
Parametrize simple boundaries when possible.
Use Lagrange multipliers on smooth regular constraint pieces.
Check endpoints, corners, and points where the constraint is not regular.
Evaluate \(f\) at every candidate and compare values for global extrema.
Worked example
Example: Find the maximum and minimum of \(f(x,y)=x\) on \(x^2+y^2=1\).
On the unit circle, the largest possible \(x\)-coordinate is \(1\) and the smallest is \(-1\). Lagrange multipliers give the same candidates: \((1,0)\) gives maximum value \(1\), and \((-1,0)\) gives minimum value \(-1\). For the related product \(xy\) on the same circle, the diagonal candidates give maximum value \(1/2\).
Try it
Try it: At a regular constrained extremum of \(f\) subject to \(g=c\), what condition should hold?
Hint: At a regular constrained extremum, the two gradients are parallel.
Most mistakes come from using the right test in the wrong setting
Learning goal: Finish with a compact workflow and avoid the common false conclusions.
Common traps
Gradient zero is not classification: it only finds candidates.
Tangent plane is local: it approximates the graph near the base point.
\(D=0\) is inconclusive: test signs or use higher-order terms.
Boundary points matter: global extrema can occur where \(\nabla f≠0\).
Compactness matters: continuous functions on noncompact sets need not attain extrema.
Example: Find global extrema of \(f(x,y)=x^2+y^2\) on the disk \(x^2+y^2\le1\).
The interior critical point is \((0,0)\), where \(f=0\), giving the global minimum. On the boundary \(x^2+y^2=1\), the function equals \(1\), so every boundary point gives the global maximum value \(1\).
Try it
Try it: If \(f\) is continuous on a compact set, what is guaranteed?
Hint: This is the extreme value theorem.
Final recap
Interior differentiable local extrema must satisfy \(\nabla f=0\).
The tangent plane to \(z=f(x,y)\) uses \(f(a,b)\), \(f_x(a,b)\), and \(f_y(a,b)\); a graph normal is \((f_x,f_y,-1)\).
In two variables, \(D=f_{xx}f_{yy}-f_{xy}^2\) classifies nondegenerate critical points.
Positive definite Hessian means strict local minimum; negative definite means strict local maximum; indefinite means saddle.
\(D=0\) or zero Hessian requires direct sign or higher-order analysis.
For constrained extrema, use \(\nabla f=\lambda\nabla g\) only on regular constraint pieces.
For global extrema on compact sets, evaluate and compare every interior, boundary, corner, and singular candidate.
Next step: Close this lesson and try the quiz again. First identify whether the question asks for a critical point, a tangent plane, a local classification, a constrained condition, or a global comparison.
Practice set
Kritieke punten, raakvlakken en lokale extrema practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
Welke voorwaarde moet gelden bij een inwendig lokaal extremum van een differentieerbare functie \(f(x,y)\)?
Correct answer: D. \(\nabla f=0\)
Explanation: Bij een inwendig extremum is de eerstordevoorwaarde \(\nabla f=0\).
Question 2Not answered
Welk type punt is \((0,0)\) voor \(f(x,y)=x^2+y^2\)?
Correct answer: A. Lokaal minimum
Explanation: De functie is niet-negatief en is alleen \(0\) in de oorsprong.
Question 3Not answered
Welk type punt is \((0,0)\) voor \(f(x,y)=x^2-y^2\)?
Correct answer: D. Zadelpunt
Explanation: De functie is positief langs de \(x\)-as en negatief langs de \(y\)-as.
Question 4Not answered
Als de Hessiaan in een kritisch punt positief-definiet is, wat suggereert dat?
Correct answer: B. Strikt lokaal minimum
Explanation: Een positief-definiete Hessiaan geeft een strikt lokaal minimum.
Question 5Not answered
Als de Hessiaan in een kritisch punt negatief-definiet is, wat suggereert dat?
Correct answer: C. Strikt lokaal maximum
Explanation: Een negatief-definiete Hessiaan geeft een strikt lokaal maximum.
Question 6Not answered
Als de Hessiaan in een kritisch punt onbepaald is, wat geeft dat meestal aan?
Correct answer: A. Zadelpunt
Explanation: Een onbepaalde Hessiaan betekent dat de functie in sommige richtingen omhoog kromt en in andere omlaag.
Question 7Not answered
Wat is het raakvlak bij \((a,b)\) voor \(z=f(x,y)\)?
Correct answer: A. \(z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\)
Explanation: De lineaire benadering gebruikt de functiewaarde en de twee eerste partiële afgeleiden.
Question 8Not answered
Bij een extremum onder een beperking van \(f\) met \(g=c\), zeggen Lagrange-vermultipliers:
Correct answer: B. \(\nabla f=\lambda\nabla g\)
Explanation: De gradiënten zijn parallel bij regelmatige extremen onder een beperking.
Question 9Not answered
Als \(f\) continu is op een compacte verzameling, dan:
Correct answer: D. Neemt een maximum en een minimum aan
Explanation: De stelling van de extreme waarden garandeert beide extrema.
Question 10Not answered
Voor \(f(x,y)=xy\), welk type punt is \((0,0)\)?
Correct answer: C. Zadelpunt
Explanation: Langs \(y=x\) is \(f=x^2\); langs \(y=-x\) is \(f=-x^2\).
