Skip to content
Calculus Solver

Critical Points Calculator

Find every critical point of a function in an interval and classify each one as a local maximum, a local minimum or neither. Roots are exact whenever the derivative factors nicely; otherwise they are found numerically and marked as approximate.

Your expression will be shown here as it will be read.

Solved entirely in your browser. Nothing you type is sent to a server or stored, and no account is needed.

Try an example

What is a critical point?

A curve with a flat tangent at the top of a hill and at the bottom of a valley, marking a local maximum and minimum

A critical point of f is a number c in the domain of f where either f′(c) = 0 (a horizontal tangent) or f′(c) does not exist (a corner, cusp or vertical tangent). Fermat's theorem says every local maximum or minimum of a function on an open interval occurs at a critical point.

The converse is false: not every critical point is an extremum. f(x) = x³ has f′(0) = 0, but the graph keeps rising through the origin. That is why critical points must be classified.

c is critical  ⟺  c∈dom⁡f and (f′(c)=0 or f′(c) undefined)c \text{ is critical} \iff c \in \operatorname{dom} f \ \text{and}\ \big(f'(c) = 0 \ \text{or}\ f'(c) \text{ undefined}\big)

How critical points are classified

Two standard tests decide what kind of point each candidate is.

  • Second derivative test

    f′(c)=0, f′′(c)>0⇒local min;f′′(c)<0⇒local maxf'(c) = 0,\ f''(c) > 0 \Rightarrow \text{local min};\quad f''(c) < 0 \Rightarrow \text{local max}

    Quick when f″ is easy to evaluate. It is inconclusive when f″(c) = 0.

  • First derivative test

    f′:−→+ ⇒min,f′:+→− ⇒maxf' : - \to + \ \Rightarrow \text{min},\qquad f' : + \to - \ \Rightarrow \text{max}

    Look at the sign of f′ just left and right of c. It also works where f′ is undefined.

  • No sign change

    f′:+→+ or −→− ⇒neitherf' : + \to + \ \text{or}\ - \to - \ \Rightarrow \text{neither}

    The graph flattens momentarily but keeps going the same way, as x³ does at 0.

How this calculator works

The calculator differentiates f with full steps, then splits f′ into a numerator and a denominator. Polynomial factors of the numerator are solved exactly, using the rational root theorem and then the quadratic formula. That is how answers like x = ±√2 appear in exact form.

Zeros of the denominator, and of any absolute value inside f′, are candidates where f′ is undefined; they count only if f itself is defined there. A dense numerical scan of the search interval then catches any remaining roots, such as those of trigonometric or exponential derivatives.

Each point is classified with the second derivative test when f″ is nonzero, and with the first derivative test otherwise. The function value at each point is computed exactly when possible.

For the full details, including how answers are verified, see how the solver works.

Worked examples

Work through these by hand first, then enter them above to compare each step.

Example 1

A cubic with a maximum and a minimum

f(x)=x3−3x2−9x+5f(x) = x^{3} - 3x^{2} - 9x + 5

  1. Differentiate and factor:

    f′(x)=3x2−6x−9=3(x−3)(x+1)f'(x) = 3x^{2} - 6x - 9 = 3(x - 3)(x + 1)

  2. The critical points are x = −1 and x = 3. Use f″(x) = 6x − 6:

    f′′(−1)=−12<0 (max),f′′(3)=12>0 (min)f''(-1) = -12 < 0\ (\text{max}),\qquad f''(3) = 12 > 0\ (\text{min})

  3. Function values: f(−1) = 10 and f(3) = −22.

Answer

x=−1 (local max, f=10),x=3 (local min, f=−22)x = -1\ (\text{local max},\ f = 10),\quad x = 3\ (\text{local min},\ f = -22)

Example 2

A quartic with three critical points

f(x)=x4−2x2f(x) = x^{4} - 2x^{2}

  1. Differentiate and factor completely:

    f′(x)=4x3−4x=4x(x−1)(x+1)f'(x) = 4x^{3} - 4x = 4x(x - 1)(x + 1)

  2. Classify with f″(x) = 12x² − 4:

    f′′(±1)=8>0,f′′(0)=−4<0f''(\pm 1) = 8 > 0,\qquad f''(0) = -4 < 0

Answer

x=±1 (local min, f=−1),x=0 (local max, f=0)x = \pm 1\ (\text{local min},\ f = -1),\quad x = 0\ (\text{local max},\ f = 0)

Example 3

A cusp where f′ is undefined

f(x)=x2/3f(x) = x^{2/3}

  1. The derivative has x in the denominator:

    f′(x)=23x−1/3=23x3f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}

  2. f′(x) is never 0, but it is undefined at x = 0, where f(0) = 0 is defined. f′ changes from negative to positive there.

Answer

x=0 (local min, a cusp)x = 0\ (\text{local min, a cusp})

Example 4

An exponential example

f(x)=x e−xf(x) = x\,e^{-x}

  1. Product rule:

    f′(x)=e−x−xe−x=e−x(1−x)f'(x) = e^{-x} - x e^{-x} = e^{-x}(1 - x)

  2. Since e^(−x) > 0 always, f′ = 0 only at x = 1, where f′ changes from + to −.

Answer

x=1 (local max, f=1e)x = 1\ \left(\text{local max},\ f = \tfrac{1}{e}\right)

Common mistakes

  • Assuming f″(c) = 0 means “neither”

    x⁴ has f″(0) = 0 yet a clear minimum at 0. When the second derivative test is inconclusive, switch to the first derivative test.

  • Forgetting points where f′ is undefined

    |x| and x^(2/3) have their minimum exactly where f′ fails to exist. Check the denominator of f′ as well as the numerator.

  • Including points outside the domain

    For f(x) = 1/x, f′ is undefined at 0, but 0 is not in the domain of f, so it is not a critical point.

  • Mixing up local and absolute extrema

    On a closed interval [a, b], the absolute maximum may occur at an endpoint. Compare the endpoint values with the critical values yourself.

Limitations

  • Only the search interval you set is examined (default −10 to 10). Periodic functions have infinitely many critical points, and only those in the interval are listed.
  • Roots of non-polynomial derivatives are found numerically and marked ≈. Very close or tangent roots can occasionally be missed; narrowing the interval helps.
  • Absolute extrema on closed intervals and endpoint analysis are not computed automatically.
  • Functions of more than one variable (saddle points, gradients) are not supported.

Related calculators

Frequently asked questions

Are critical points the same as local maxima and minima?

No. Every local extremum (in the interior of the domain) is a critical point, but some critical points are neither, like x = 0 for x³. Classification decides which is which.

What if the second derivative is zero?

The second derivative test is then inconclusive, so the calculator falls back to the first derivative test, checking the sign of f′ on each side of the point.

Why do some answers show ≈?

When the derivative does not factor into polynomials, the critical points are found numerically to about 10 significant digits and marked with ≈ so you know they are approximations.

Can I change where the calculator searches?

Yes. Edit the “Search from” and “Search to” boxes. Narrow intervals give the numerical search more precision; wide ones help when you don't know where the points are.

Next step

Pick one of the worked examples, solve it on paper, then enter it in the calculator and compare your steps with each rule shown. Spotting where your work differs is the fastest way to improve.

Back to the calculator