How to Find Critical Points (and Classify Them) Step by Step

You can find critical points. First, calculate a function’s derivative, f'(x). Then set f'(x) = 0 to solve for x. Also, identify where f'(x) is undefined. However, f(x) must remain defined at these points. To classify each critical point, use the First Derivative Test. This tests the sign changes of f'(x) across intervals. Alternatively, you can use the Second Derivative Test. Check the sign of the second derivative f”(c) with this method. You will classify each point as a local maximum, local minimum, or neither.

A critical point (or critical number c) of a function f(x) is an x-value in the domain of f where either f'(c) = 0 (a horizontal tangent) or f'(c) is undefined (a corner, cusp, or vertical tangent). Every local peak or valley of a smooth function must occur at a critical point (Fermat’s Theorem), making critical points the foundation for solving calculus extrema and optimization problems.

Why Fermat’s Theorem guarantees local extrema occur at critical points

Picture the top of a smooth hill on a graph. Just before reaching the peak, the function rises. This means tangent lines have positive slopes (f'(x) > 0). Just after the peak, the function falls. This means tangent lines have negative slopes (f'(x) < 0). The derivative of a continuous, differentiable function changes continuously. So, the slope must pass through exactly zero at the precise peak, which explains how to find critical points. A similar mirror-image behavior happens at the bottom of a valley.

Mathematically, suppose f has a local maximum at c and f'(c) exists. For a small positive step h > 0, f(c + h) ≤ f(c), which makes the difference quotient (f(c + h) – f(c)) / h less than or equal to 0. For a small negative step h < 0, the numerator remains non-positive while the denominator is negative, making the difference quotient greater than or equal to 0. Taking the limit as h approaches 0 from both sides forces f'(c) to be simultaneously ≤ 0 and ≥ 0, which means f'(c) = 0.

Consider a sharp bend, like the point (0, 0) on f(x) = |x|. A peak or valley occurs here. However, the left-hand derivative (-1) does not equal the right-hand derivative (+1). Because of this, the derivative does not exist at that point. This is why the definition for how to find critical points explicitly includes values where f'(x) is undefined.

The converse is not true: finding f'(c) = 0 does not guarantee that c is a maximum or minimum. Critical points are potential candidates, and you must classify them using derivative tests.

a step-by-step guide on how to find critical points

Follow these four steps for any algebraic, trigonometric, exponential, or logarithmic function:

1. Find the domain of f(x). Note any x-values where the original function is undefined. For example, watch for division by zero. Also, look for negative values under even roots. Critical points can only exist inside the domain of f.
2. Calculate the derivative f'(x). Use derivative techniques such as the power rule, product rule, or quotient rule. You can use a derivative calculator to verify your derivative steps for complex functions.
3. Find all critical numbers. Set f'(x) = 0 and solve for x. Then determine where f'(x) is undefined while checking that f(x) itself exists at those values.
4. Classify each critical point. You can use either the First Derivative Test or the Second Derivative Test. This will determine if each critical number represents a local maximum, local minimum, or neither. Calculate f(c) to state the full coordinate (c, f(c)).

Comparing the First and Second Derivative Tests

Once you locate critical numbers, you must determine what is happening on the graph at those locations. You have two testing methods available, each with specific advantages and drawbacks.

The First Derivative Test

The First Derivative Test evaluates the sign of f'(x). It checks open intervals immediately to the left and right of a critical number c. You will look for a positive or negative sign.

• Local maximum: f'(x) changes from positive (+) to negative (-) as x moves from left to right through c. The graph increases then decreases.
• Local minimum: f'(x) changes from negative (-) to positive (+) as x moves from left to right through c. The graph decreases then increases.
• Neither (inflection point): f'(x) retains the same sign on both sides of c (either + to + or – to -). The graph flattens temporarily but continues in the same vertical direction.

The Second Derivative Test

The Second Derivative Test evaluates concavity at a critical point. Here f'(c) = 0. You can plug c directly into the second derivative f”(x).

• If f”(c) > 0, the graph is concave upward (shaped like a bowl), meaning c is a local minimum.
• Next, if f”(c) < 0, the graph is concave downward (shaped like an umbrella), meaning c is a local maximum.
• If f”(c) = 0, the test is inconclusive. The test is also inconclusive if f”(c) is undefined. You must switch to the First Derivative Test.

Feature	First Derivative Test	Second Derivative Test
Applicability	Works on all critical points (f'(c) = 0 or f'(c) undefined)	Only works when f'(c) = 0 and f”(c) exists
Method	Tests signs of f'(x) on test points in nearby intervals	Evaluates f”(c) directly at the critical value c
Inconclusive outcomes	Never fails; always yields a definitive answer	Fails when f”(c) = 0 (requires First Derivative Test)
Best used for	Functions with undefined derivatives or complex sign charts	Polynomially simple functions where f”(x) is easy to compute

Worked examples across different function types

Example 1: Polynomial function (cubic)

Find and classify the critical points of f(x) = 2x^3 – 3x^2 – 12x + 5.

Step 1: Derivative.
f'(x) = 6x^2 – 6x – 12 = 6(x^2 – x – 2) = 6(x – 2)(x + 1)

Step 2: Solve f'(x) = 0.
First, 6(x – 2)(x + 1) = 0. This gives critical numbers x = -1 and x = 2. The derivative is a polynomial, so f'(x) is defined everywhere.

Step 3: Classify using the Second Derivative Test.
f”(x) = 12x – 6

• At x = -1: f”(-1) = 12(-1) – 6 = -18 < 0. Because concavity is downward, x = -1 is a local maximum. Evaluating f(-1) gives f(-1) = 2(-1)^3 – 3(-1)^2 – 12(-1) + 5 = 12. The local maximum point is (-1, 12).
• At x = 2: f”(2) = 12(2) – 6 = 18 > 0. Because concavity is upward, x = 2 is a local minimum. Evaluating f(2) gives f(2) = 2(8) – 3(4) – 12(2) + 5 = -15. The local minimum point is (2, -15).

Example 2: When the Second Derivative Test fails

Find and classify the critical points of f(x) = x^4 – 4x^3.

Step 1: Derivative.
f'(x) = 4x^3 – 12x^2 = 4x^2(x – 3)

Setting f'(x) = 0 yields critical values x = 0 and x = 3.

Step 2: Second derivative check.
f”(x) = 12x^2 – 24x

• At x = 0: f”(0) = 0. The Second Derivative Test fails completely here.
• At x = 3: f”(3) = 12(9) – 24(3) = 36 > 0, which indicates a local minimum at (3, -27).

Step 3: First Derivative Test for x = 0.
Test x in intervals around 0. For example, if x < 0, use x = -1. Then f'(-1) = 4(-1)^2(-1 – 3) = -16 < 0. Also, for 0 < x < 3, use x = 1. Then f'(1) = 4(1)^2(1 – 3) = -8 < 0. The sign of f'(x) does not change. It remains negative on both sides of 0. Therefore, x = 0 is neither a maximum nor a minimum, for how to find critical points in this function. It represents a horizontal inflection point.

Example 3: Function with an undefined derivative

Find and classify the critical points of f(x) = x^(2/3).

Step 1: Derivative.
f'(x) = (2/3) x^(-1/3) = 2 / (3 x^(1/3))

Step 2: Critical points.
The numerator is always 2. Therefore, the equation f'(x) = 0 has no solutions. However, f'(x) is undefined at x = 0. This is because of division by zero. Since f(0) = 0 exists in the original domain, x = 0 is a critical point for how to find critical points.

Step 3: Classify using the First Derivative Test.
For x < 0, f'(x) is negative because x^(1/3) is negative. For x > 0, f'(x) is positive. Because f'(x) changes from negative to positive at x = 0, the point (0, 0) is a local minimum (specifically a sharp cusp).

Example 4: Trigonometric function on a restricted interval

Find all critical points of f(x) = sin x + cos x on the interval [0, 2π].

Step 1: Derivative.
f'(x) = cos x – sin x

Step 2: Solve f'(x) = 0.
cos x – sin x = 0 implies sin x = cos x, or tan x = 1. On [0, 2π], tan x = 1 at x = π/4 and x = 5π/4.

Step 3: Classify using the Second Derivative Test.
f”(x) = -sin x – cos x = -f(x)

• At x = π/4: f(π/4) = √2 / 2 + √2 / 2 = √2 > 0, so f”(π/4) = -√2 < 0. This is a local maximum with value f(π/4) = √2.
• At x = 5π/4: f(5π/4) = -√2 / 2 – √2 / 2 = -√2 < 0, so f”(5π/4) = √2 > 0. This is a local minimum with value f(5π/4) = -√2.

Example 5: Rational function

Find the critical points of f(x) = x / (x^2 + 1).

Applying the quotient rule yields:

f'(x) = [(x^2 + 1)(1) – x(2x)] / (x^2 + 1)^2 = (1 – x^2) / (x^2 + 1)^2

The denominator (x^2 + 1)^2 is always greater than or equal to 1, so f'(x) is defined everywhere. Setting the numerator to zero gives 1 – x^2 = 0, so x = -1 and x = 1.

The denominator is always positive, so the sign of f'(x) depends entirely on 1 – x^2:

• For x < -1, f'(x) < 0.
• Also, for -1 < x < 1, f'(x) > 0.
• For x > 1, f'(x) < 0.

Therefore, x = -1 is a local minimum with f(-1) = -1/2, and x = 1 is a local maximum with f(1) = 1/2.

Example 6: Logarithmic function and domain boundaries

Find the critical points of f(x) = x ln x.

The domain of f(x) requires x > 0. Using the product rule gives:

f'(x) = (1)(ln x) + x(1/x) = ln x + 1

Setting f'(x) = 0 gives ln x = -1, so x = e^(-1) = 1/e. For 0 < x < 1/e, f'(x) < 0; for x > 1/e, f'(x) > 0. Thus, x = 1/e is a local minimum with value f(1/e) = -1/e.

Now consider g(x) = 1/x. Its derivative g'(x) = -1/x^2 is undefined at x = 0. However, x = 0 is not in the domain of g(x). Therefore, x = 0 is not a critical point. For how to find critical points, remember they can only exist where the function itself is defined.

Finding absolute extrema on a closed interval

When you find absolute (global) maximum and minimum values, use the Closed Interval Method. This applies to a continuous function on a closed interval [a, b].

1. Find all critical numbers of f in the open interval (a, b).
2. Calculate f(c) for each critical number c.
3. Calculate the endpoint values f(a) and f(b).
4. Compare all calculated values: the largest output is the absolute maximum, and the smallest output is the absolute minimum.

For example, to find the absolute extrema of f(x) = x^3 – 3x on [-3, 3]:

f'(x) = 3x^2 – 3 = 3(x – 1)(x + 1), giving critical numbers x = -1 and x = 1, both inside [-3, 3]. Evaluate f at all candidate values:

• f(-3) = (-3)^3 – 3(-3) = -18 (endpoint)
• f(-1) = (-1)^3 – 3(-1) = 2 (critical point)
• f(1) = (1)^3 – 3(1) = -2 (critical point)
• f(3) = (3)^3 – 3(3) = 18 (endpoint)

Comparing these outputs reveals an absolute maximum of 18 at x = 3 and an absolute minimum of -18 at x = -3.

Common mistakes when finding critical points

• Dividing away variable factors: Solving 4x^3 = 12x^2 by dividing both sides by x^2 loses the critical value x = 0. Always factor equations (e.g., 4x^2(x – 3) = 0) to avoid losing valid solutions.
• Ignoring undefined derivative points: You might forget to check where denominators or fractional exponents equal zero. As a result, you will omit essential critical numbers. These include cusps and vertical tangents.
• Including numbers outside the function domain: Values where f(x) itself is undefined (such as vertical asymptotes like x = 0 for 1/x) cannot be critical points.
• Mistaking f”(c) = 0 for “neither”: A second derivative of zero means the Second Derivative Test fails, not that there is no extrema. Functions like f(x) = x^4 have f”(0) = 0 but still possess a local minimum at x = 0.
• Reporting only the x-coordinate: In multi-variable problems, sketch tasks, or formal exam questions, a point requires both coordinates (c, f(c)). Always evaluate f(c) in the original function.

Practical applications of critical points

Finding critical points is crucial across numerous scientific and technical disciplines:

• Calculus optimization: Solving real-world optimization tasks such as minimizing material cost for containers or maximizing physical enclosures.
• Curve sketching and analysis: Determining increasing/decreasing intervals and mapping turning points using the Mean Value Theorem and Rolle’s Theorem.
• Economics and finance: Locating peak profit margins by setting marginal revenue equal to marginal cost.
• Physics and engineering: Identifying stable and unstable physical equilibrium states where potential energy derivatives equal zero.

Critical points calculator gadget

Enter a function and an interval below to solve f'(x) = 0, classify each point, and output absolute extrema:

For additional space and complex input controls, visit the full critical points calculator page.

Practice problems with solutions

Solve these problems independently, then check your steps below:

1. f(x) = x^3 – 12x
2. f(x) = x e^(-x)
3. f(x) = x^4 – 2x^2
4. f(x) = x + 4/x
5. f(x) = |x – 1|

Worked solutions

1. f(x) = x^3 – 12x:
f'(x) = 3x^2 – 12 = 3(x^2 – 4) = 3(x – 2)(x + 2). Setting f'(x) = 0 gives critical numbers x = -2 and x = 2. f”(x) = 6x. f”(-2) = -12 < 0 (local maximum at x = -2, f = 16); f”(2) = 12 > 0 (local minimum at x = 2, f = -16).

2. f(x) = x e^(-x):
f'(x) = e^(-x) + x(-e^(-x)) = e^(-x)(1 – x). To find critical points, setting f'(x) = 0 yields x = 1, since e^(-x) is never zero. For x < 1, f'(x) > 0; for x > 1, f'(x) < 0. Local maximum at x = 1 with f(1) = 1/e.

3. f(x) = x^4 – 2x^2:
f'(x) = 4x^3 – 4x = 4x(x^2 – 1) = 4x(x – 1)(x + 1). For how to find critical points, numbers are x = 0, x = -1, x = 1. f”(x) = 12x^2 – 4. f”(0) = -4 < 0 (local maximum at x = 0, f = 0); f”(±1) = 8 > 0 (local minima at x = -1 and x = 1, f = -1).

4. f(x) = x + 4/x:
f'(x) = 1 – 4/x^2. This also equals (x^2 – 4)/x^2. Set the numerator to 0. This gives x = -2 and x = 2. At x = 0, f'(x) is undefined. However, x = 0 is not in the domain of f(x). So, x = 0 is not a critical point. Now, test the signs. You will find x = -2 is a local maximum. Here, f(-2) = -4. Also, x = 2 is a local minimum. Here, f(2) = 4. Notice that a local maximum value of -4 is lower than a local minimum value of 4. This happens due to asymptotic behavior between them.

5. f(x) = |x – 1|:
Rewriting as a piecewise function, f'(x) = -1 for x < 1 and f'(x) = 1 for x > 1. At x = 1, f'(1) does not exist (corner point). Since f(1) = 0 is defined, x = 1 is a critical point. f'(x) changes from negative to positive at x = 1, so x = 1 is a local minimum.

Frequently asked questions

Is every critical point a local maximum or local minimum?

No. A critical point can be an inflection point where the derivative equals zero but does not change sign on either side, such as x = 0 for the function f(x) = x^3.

Are interval endpoints classified as critical points?

No, critical points must occur in the interior of a function’s domain. You can analyze two-sided derivative behavior there. However, you evaluate endpoints alongside critical points. This happens when you find absolute extrema on closed intervals.

What is the difference between a critical number and a critical point?

A critical number refers specifically to the x-value c where derivative criteria are met, whereas a critical point refers to the full ordered pair coordinate (c, f(c)) plotted on the graph.

Can a smooth continuous function have zero critical points?

Yes. The exponential function f(x) = e^x has derivative f'(x) = e^x, which is never zero and defined everywhere on negative infinity to positive infinity, meaning it has no critical points and no local extrema.

Sometimes, f'(x) = 0 cannot be factored algebraically. When this happens, apply Newton’s method to approximate roots. Also, explore our detailed guide on points and inflection points. This helps when analyzing higher-order curvature changes.

Further reading

• Critical Points (Paul’s Online Math Notes) — detailed worked examples across polynomial, rational, root, trigonometric, exponential, and logarithmic functions.
• Finding Absolute Extrema (Paul’s Online Math Notes) — step-by-step application of the closed interval method and avoiding common analytical traps.
• The Shape of a Graph, Part I (Paul’s Online Math Notes) — interpreting increasing and decreasing intervals alongside the First Derivative Test.