What's the difference between a critical point and a critical value?

A critical point is the x-coordinate c where f'(c) = 0 or f'(c) is undefined. A critical value is the y-coordinate, f(c). When finding critical points, always report the x-values; critical values are the corresponding y-outputs if needed later.

Can a critical point be at an endpoint of the domain?

By the definition of critical points, no—they occur where the derivative is zero or undefined within the interior of the domain. However, endpoints are tested separately when finding absolute extrema, even though they aren't technically critical points.

What if the derivative is never zero?

Then the function has no critical points (at least from f'(c) = 0). Check whether the derivative is undefined anywhere. If the derivative exists and is never zero, the function is strictly increasing or decreasing everywhere.

How many critical points can a function have?

There's no upper limit—a polynomial of degree n can have up to n-1 critical points because its derivative has degree n-1. Transcendental functions like \sin(x) have infinitely many critical points.

Critical Points

Critical points mark where a function's derivative is zero or undefined, revealing local extrema and where the graph changes direction.

f'(c) = 0 \text{ or } f'(c) \text{ undefined}

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Type the problem. The solver will use Critical Points where Critical Points is the right tool, and tell you when it is not.

What each symbol means

What Critical Points takes

Symbol	Meaning
$f$	The function being analyzed; if you misread this as a variable or number, you'll compute derivatives of the wrong thing.
$c$	The x-coordinate of a critical point where the derivative condition holds; it's not a slope or a function value, but an input value.

When to use it

Reach for critical points when you need to find local maxima and minima, or understand where a function's behavior shifts.

Level

Usually taught in: Calculus I · Appears on: AP Calculus

Worked examples

1. Find critical points of a cubic polynomial

Problem

Find the critical points of

f(x) = x^3 - 3x^2

$f'(x) = 3x^2 - 6x$
Differentiate using the power rule, reducing each exponent by 1 and multiplying by the original exponent.
$3x^2 - 6x = 0$
Set the derivative equal to zero to find where the slope is zero.
$3x(x - 2) = 0$
Factor out the common factor of $3x$ .
$x = 0 \text{ or } x = 2$
Each factor equals zero, giving us the two critical points.

Answer: $x = 0 \text{ or } x = 2$

We use the power rule to find the derivative, set it equal to zero, and factor to find the values of $x$ where the slope is zero. These are critical points because the derivative changes sign at these locations.

2. Find critical points of a rational function

Problem

Find the critical points of

f(x) = \frac{x}{x^2 + 1}

$f'(x) = \frac{(x^2 + 1) \cdot 1 - x \cdot 2x}{(x^2 + 1)^2}$
Apply the quotient rule: if $f(x) = \frac{u}{v}$ , then $f'(x) = \frac{vu' - uv'}{v^2}$ .
$f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2}$
Multiply out the numerator: $(x^2 + 1) \cdot 1 = x^2 + 1$ and $x \cdot 2x = 2x^2$ .
$f'(x) = \frac{1 - x^2}{(x^2 + 1)^2}$
Combine like terms in the numerator: $x^2 + 1 - 2x^2 = 1 - x^2$ .
$1 - x^2 = 0$
Set the numerator equal to zero; the denominator $(x^2 + 1)^2$ is always positive.
$x^2 = 1$
Add $x^2$ to both sides.
$x = -1 \text{ or } x = 1$
Take the square root of both sides to get both the negative and positive square root.

Answer: $x = -1 \text{ or } x = 1$

The quotient rule produces a more complex derivative. Critical points occur where either the numerator is zero OR the denominator is undefined. Here, the denominator is always positive, so we only set the numerator equal to zero, yielding $x = -1$ and $x = 1$ as critical points.

3. Optimize pricing tiers in a data plan

Problem

A phone plan charges forty dollars per month plus two dollars per gigabyte of data. A company's monthly profit (in thousands of dollars) from their pricing tier structure is

P(x) = -x^3 + 12x^2 - 36x + 100

, where

x

is the number of tiers offered (between 1 and 10). Find the critical points.

$P'(x) = -3x^2 + 24x - 36$
Differentiate the profit function using the power rule on each term.
$-3x^2 + 24x - 36 = 0$
Set the derivative equal to zero to find where the rate of change of profit is zero.
$-3(x^2 - 8x + 12) = 0$
Factor out the common factor $-3$ .
$x^2 - 8x + 12 = 0$
Divide both sides by $-3$ .
$(x - 2)(x - 6) = 0$
Factor the quadratic: we need two numbers that multiply to 12 and add to $-8$ , which are $-2$ and $-6$ .
$x = 2 \text{ or } x = 6$
Each factor equals zero, giving the two critical points where profit reaches local extrema.

Answer: $x = 2 \text{ or } x = 6$

Critical points reveal where profit reaches local maxima or minima. By differentiating the profit function and setting it equal to zero, we find that offering either 2 or 6 different pricing tiers yields critical points. The company would use the first or second derivative test to determine which tier count produces a local maximum profit.

Common mistakes

Where Critical Points usually goes wrong

The mistake	Why it is wrong	The fix
Setting $f'(c) = 0$ but forgetting to check where $f'(c)$ is undefined.	Functions can have critical points where the derivative doesn't exist (like at sharp corners or vertical tangents), and missing these means you've found only a partial list.	After finding where $f'(c) = 0$ , check the domain and look for points where the derivative involves division by zero, absolute values, or fractional powers that become undefined.
Finding $f'(x) = 0$ correctly but computing $f'(x)$ incorrectly using wrong derivative rules.	If you misapply the quotient rule, product rule, or chain rule when finding the derivative, you'll get a wrong derivative and solve the wrong equation.	Double-check each derivative calculation, especially with products, quotients, and composite functions—verify using alternate methods or careful rule application.
Confusing critical points with critical values—writing down $f(c)$ instead of $c$ as the answer.	A critical point is the x-coordinate (the input) where $f'(c) = 0$ or is undefined, not the y-coordinate; reporting $f(c)$ doesn't tell you where the derivative is zero.	Always report the x-values (the numbers $c$ satisfying the condition), not the y-values; if asked for a critical point as an ordered pair, write $(c, f(c))$ .

Tips and when to use something else

After finding critical points, use the First Derivative Test or Second Derivative Test to determine whether each is a local maximum, local minimum, or neither.
Watch for points where the derivative is undefined due to a discontinuity or sharp corner in the function—these count as critical points too.
Not every critical point is a local extremum; some functions have horizontal tangents (where $f'(c) = 0$ ) at inflection points instead.
If the derivative is a rational function, factor the numerator and denominator separately to find all critical points; don't forget roots of the denominator.

Related formulas

First Derivative TestClassify critical points as local maxima or minima by checking whether the derivative changes sign before and after the point.Second Derivative TestThe Second Derivative Test determines whether a critical point is a local minimum or maximum using the concavity at that point.Definition of the DerivativeThe Definition of the Derivative finds the instantaneous rate of change at a point using limits; the foundation for all derivative calculations.Power RuleThe Power Rule states that the derivative of x^n is nx^{n-1}, making it the essential tool for differentiating any polynomial expression.Concavity and Inflection PointsPinpoint where a function's curve changes from bending upward to downward, revealing the transition point between concave up and concave down.

Frequently asked questions

What's the difference between a critical point and a critical value?: A critical point is the x-coordinate $c$ where $f'(c) = 0$ or $f'(c)$ is undefined. A critical value is the y-coordinate, $f(c)$ . When finding critical points, always report the x-values; critical values are the corresponding y-outputs if needed later.
Can a critical point be at an endpoint of the domain?: By the definition of critical points, no—they occur where the derivative is zero or undefined within the interior of the domain. However, endpoints are tested separately when finding absolute extrema, even though they aren't technically critical points.
What if the derivative is never zero?: Then the function has no critical points (at least from $f'(c) = 0$ ). Check whether the derivative is undefined anywhere. If the derivative exists and is never zero, the function is strictly increasing or decreasing everywhere.
How many critical points can a function have?: There's no upper limit—a polynomial of degree $n$ can have up to $n-1$ critical points because its derivative has degree $n-1$ . Transcendental functions like $\sin(x)$ have infinitely many critical points.

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Reviewed 2026-09-18