Critical Points Calculator

The critical points calculator finds the x values where a function's derivative equals zero or is undefined, identifying the candidates for local maxima, minima, and saddle points. It is used by calculus students, economists optimising functions, and engineers finding peak or trough values. Enter your function to locate all critical points with full step-by-step working.

How to Use the Critical Points Calculator

1. Enter your function f(x) in the Critical Points Calculator input field. Click Calculate to find the derivative f'(x) automatically.
2. Review the derivative, then see the solutions to f'(x) = 0 listed as critical points.
3. Check whether the derivative is undefined at any additional x values, which are also critical points.
4. Use the second derivative test results shown to classify each point as a local max, min, or neither.

The Formula

A critical point occurs where:

f'(x) = 0, or f'(x) is undefined

To find critical points:

1. Differentiate f(x) to get f'(x)
2. Solve f'(x) = 0 for x
3. Identify any x values where f'(x) does not exist (e.g., at corners, cusps, or discontinuities)

To classify each critical point, apply the second derivative test:

• If f''(x) > 0: local minimum
• If f''(x) < 0: local maximum
• If f''(x) = 0: the test is inconclusive; use the first derivative sign test instead

Real-World Example

Find and classify the critical points of f(x) = x^3 - 6x^2 + 9x + 1.

1. Differentiate: f'(x) = 3x^2 - 12x + 9
2. Set f'(x) = 0: 3x^2 - 12x + 9 = 0, which gives x^2 - 4x + 3 = 0
3. Factorise: (x - 1)(x - 3) = 0, so x = 1 and x = 3
4. Find f''(x) = 6x - 12
5. At x = 1: f''(1) = -6 < 0, so x = 1 is a local maximum
6. At x = 3: f''(3) = 6 > 0, so x = 3 is a local minimum

The function values are f(1) = 5 (local max) and f(3) = 1 (local min).

Using Critical Points in Optimisation

Critical points are the key to finding the best or worst values of a function within a domain. In business, they locate the production level that maximises profit or minimises cost. In engineering, they identify the angle that maximises beam strength or the shape that minimises material use. To find the absolute maximum or minimum over a closed interval, evaluate f(x) at each critical point inside the interval and also at both endpoints, then compare all values. The largest is the absolute maximum, the smallest the absolute minimum.

Frequently Asked Questions

Are all critical points local maxima or minima? No. A critical point can also be a point of inflection where the derivative is zero but the function does not change direction. For example, f(x) = x^3 at x = 0 has f'(0) = 0, but the function continues increasing on both sides, making it a stationary point of inflection rather than an extremum.

What is the first derivative test? Check the sign of f'(x) on each side of the critical point. If f' changes from positive to negative, the point is a local maximum. If f' changes from negative to positive, it is a local minimum. If the sign does not change, the point is neither.

Can a function have infinitely many critical points? Yes. For example, f(x) = sin(x) has critical points at x = pi/2 + n*pi for every integer n, giving infinitely many alternating maxima and minima.

How do critical points relate to inflection points? They are different concepts. Critical points occur where f'(x) = 0 or is undefined. Inflection points occur where f''(x) = 0 and the concavity changes. A point can be both simultaneously, but most critical points are not inflection points and most inflection points are not critical points.

Also try these free tools:

• Inflection Points Calculator
• Derivative Calculator
• Concavity Calculator

A second function with three critical points

The example above is a cubic with two critical points. Even-powered functions behave differently, and a quartic shows why the count of critical points is not fixed.

Take f(x) = x^4 minus 4x^2. The derivative is 4x^3 minus 8x, which factors as 4x times (x^2 minus 2). Setting the derivative to zero gives x = 0 and x = plus or minus the square root of 2, which is plus or minus 1.414214.

• At x = 0: the second derivative is 12x^2 minus 8, which gives minus 8. A negative value marks a local maximum, and f(0) is 0.
• At x = 1.414214: the second derivative is 12 times 2 minus 8, which gives 16. A positive value marks a local minimum, and f(1.414214) is 4 minus 8, which is minus 4.
• At x = minus 1.414214: the second derivative is also 16, so this is a second local minimum, with the same value of minus 4.

This function has three critical points rather than two, and two of them share the same value. The points at plus or minus 1.414214 are not the rounded values of a single point; they are two distinct solutions of x^2 = 2, and the symmetry is exact because the function contains only even powers of x. An irrational critical point is normal, and the calculator returns it to the precision it can express rather than forcing it to a fraction.

A function whose derivative is undefined

The formula for a critical point covers two cases: the derivative equal to zero, and the derivative undefined. The second case needs care, because an undefined derivative is only a critical point if the function itself is defined there.

Take f(x) = x plus 4 divided by x. The derivative is 1 minus 4 divided by x squared. Setting that to zero gives x^2 = 4, so x = 2 and x = minus 2.

• At x = 2: the second derivative is 8 divided by x cubed, which is 1. Positive, so x = 2 is a local minimum with f(2) = 4.
• At x = minus 2: the second derivative is minus 1. Negative, so x = minus 2 is a local maximum with f(minus 2) = minus 4.
• At x = 0: the derivative does not exist. The function does not exist there either, because division by zero is undefined, so x = 0 is not a critical point at all. It is a point excluded from the domain.

That last line is the one most often missed. A break in the derivative only creates a critical point if the function continues through it. An asymptote, a jump, or a removable gap in the domain can all make the derivative undefined without creating a candidate for a maximum or a minimum.

The two branches never meet. The left branch approaches the vertical axis from below, the right branch from above, and both critical points sit on their own branch. A graph drawn without the gap invites the reader to connect the two branches, which would show a maximum that does not exist.

The three functions in one table

Three functions, three different structures: two distinct extrema, three extrema with two sharing a value, and two extrema either side of a domain gap. The count of critical points follows from the degree of the derivative and the domain of the function, not from any rule about how many extrema look reasonable.

How the coefficients move the critical points

For a cubic, the position of the critical points depends on the coefficients, and the count can change. Take f(x) = x^3 minus 6x^2 plus kx plus 1. The derivative is 3x^2 minus 12x plus k, a quadratic whose roots are the critical points. The discriminant, 144 minus 12k, decides how many real roots exist.

The row for k = 12 is the instructive one. The derivative is 3 times (x minus 2) squared, which is zero at x = 2 and positive on both sides. The curve flattens at that point and then continues upward, so the point is a critical point and a stationary point of inflection, not an extremum. The second derivative there is zero as well, which means the second derivative test returns no answer and the sign of the first derivative has to settle the question.

The row for k = 13 completes the picture. The derivative is always positive, so the function increases everywhere and has no critical points at all. A function without a single critical point is perfectly ordinary, and a request for critical points should return an empty list rather than a fabricated one.

Absolute extrema on a closed interval

Critical points locate interior candidates. When the question asks for the absolute maximum or minimum on a closed interval, the endpoints are candidates too, and they are easy to forget.

On the interval from x = minus 1 to x = 4, the cubic from the first example gives four candidates.

• Left endpoint: f(minus 1) = minus 15.
• Critical point x = 1: f(1) = 5, a local maximum.
• Critical point x = 3: f(3) = 1, a local minimum.
• Right endpoint: f(4) = 5.

The absolute maximum on the interval is 5, and it occurs at both x = 1 and x = 4. The absolute minimum is minus 15 at the left endpoint, which is not a critical point. Skipping the endpoints would have named 1 as the minimum, and that answer is wrong on the interval even though it is right for the whole real line. Check every candidate before choosing, and expect ties.

How the classifications are made

Six conventions sit behind the results above.

• Critical points are the real solutions of f'(x) = 0, together with any point in the domain where f'(x) does not exist. Complex roots are ignored because they do not sit on the real line.
• A point where the derivative fails to exist counts only when the function is defined there. Domain gaps and vertical asymptotes produce discontinuities in the derivative, not candidates for extrema.
• The second derivative test classifies a point as a maximum when f'' is negative and a minimum when f'' is positive. When f'' is zero the test gives no answer, and the sign of f' either side of the point has to decide.
• A critical point whose derivative changes sign is an extremum. A critical point whose derivative does not change sign is a stationary point of inflection, and it is still reported as a critical point.
• On a closed interval, the endpoints are candidates alongside the interior critical points, and the comparison includes all of them.
• Results are returned to the precision the input supports. An irrational critical point such as the square root of 2 is reported as a decimal rather than being rounded to a convenient fraction.

One further note on domains. A function defined on a restricted domain can have an extreme value at a point that is not a critical point at all, because the restriction itself creates the boundary. The derivative knows nothing about a boundary that was imposed on the function from outside, which is why the endpoint check in the previous section is not optional.

The tests behind the classification

The classification of a local extremum by the sign of the first derivative, and by the second derivative where it is non-zero, is the standard treatment in calculus texts such as James Stewart, Calculus, in the chapters on maximum and minimum values and on curve sketching. The underlying result is Fermat's theorem: if a function has a local extremum at an interior point and the derivative exists there, then the derivative is zero at that point. That theorem is why the search for extrema starts with the derivative, and it is also why it does not finish there, since it says nothing about endpoints or about points where the derivative fails to exist.