Critical Points Calculator

Find every x where the slope is zero, then classify each as a local maximum, minimum or neither, with a sign chart for f′ and the absolute extrema on your interval.

Reads as x3−6x2+9x+1x^{3} - 6 x^{2} + 9 x + 1
Try:

Critical points

  • x=1x = 1 local maximum
  • x=3x = 3 local minimum
-112345-101020
  • f
  • f′
f′ > 0 ↗f′ < 0 ↘f′ > 0 ↗

Step by step

  1. Differentiate

    f′(x)=3x2−12x+9f'(x) = 3 x^{2} - 12 x + 9
  2. Solve f′(x) = 0

    Searching on [-1, 5].

    x=1,x=3x = 1,\quad x = 3
  3. Classify x = 1

    First derivative test: local maximum. Second derivative test: f″ = -6 < 0 → max.

    f(1)=5f(1) = 5
  4. Classify x = 3

    First derivative test: local minimum. Second derivative test: f″ = 6 > 0 → min.

    f(3)=1f(3) = 1
Closed interval method
xxtypef(x)f(x)
-1endpoint · absolute min-15
1critical5
3critical1
5endpoint · absolute max21

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An AI tutor reads the problem and steps above and explains them in plain words. It can make mistakes, so check it against the worked steps.

Finding and classifying critical points

  1. Differentiate: find f′(x)f'(x) (see the derivative calculator).
  2. Solve f′(x)=0f'(x) = 0, and note where f′f' is undefined but ff is defined.
  3. Make a sign chart for f′f'. Where the sign flips from + to −, the graph goes up then down: a local maximum. From − to +: a local minimum.
  4. Or use the second derivative: f′′(c)>0f''(c) > 0 means the curve is cupped upward there (a minimum).

A critical point with no sign change, like x=0x = 0 for x3x^3, is neither. It often turns out to be an inflection point instead.

Questions students ask

What is a critical point?

A point in the domain where f′(x)=0f'(x) = 0 or f′(x)f'(x) does not exist. Every local maximum or minimum of a differentiable function happens at a critical point, but not every critical point is an extremum.

How do I tell a maximum from a minimum?

First derivative test: if f′f' changes from + to − it is a local max; from − to + a local min; no sign change means neither. Second derivative test: f′′(c)<0f''(c) < 0 means max, f′′(c)>0f''(c) > 0 means min, and f′′(c)=0f''(c) = 0 is inconclusive.

What about absolute (global) extrema on a closed interval?

Evaluate ff at every critical point inside [a,b][a, b] and at both endpoints. The largest value is the absolute maximum and the smallest is the absolute minimum (the closed interval method). The table below does exactly this.

Why does it search only in a window?

Critical points are found numerically on the interval you choose and then snapped to exact values where possible. Functions like sin⁡x\sin x have infinitely many, so a finite window is needed.