Key takeaways
- A critical point is where f′(x) = 0 (or where f′ does not exist).
- For a cubic ax³ + bx² + cx, solve the quadratic 3ax² + 2bx + c = 0.
- Use the second derivative: f″ < 0 means a maximum, f″ > 0 means a minimum.
- The #1 mistake is solving f(x) = 0 instead of f′(x) = 0.
In this guide
What is a critical point?
A derivative gives the slope of a graph. At the top of a hill or the bottom of a valley, the graph is flat for a moment, so the slope is 0. A critical point is any x where f′(x) = 0 (or where the derivative does not exist).
A critical point is a candidate for a local maximum (a hilltop) or a local minimum (a valley floor). Sometimes it is neither, just a flat step on the way up, like x³ at x = 0.
Why do critical points matter?
Whenever you want the best of something (most profit, least material, shortest time), the answer is usually at a critical point. That is how calculus solves real optimisation problems.
A graph can only turn around where it is flat (or has a sharp corner). So to find every hilltop and valley, you only need to check a few special points, not every x.
The formula
This calculator works with cubic functions without a constant term:
| Symbol | What it means | Example |
|---|---|---|
| a | The number in front of x³ | 1 |
| b | The number in front of x² | −6 |
| c | The number in front of x | 9 |
| f′(x) | The derivative, found with the power rule | 3x² − 12x + 9 |
| x | The critical points, where f′(x) = 0 | 1 and 3 |
The derivative comes from the power rule, and setting it to zero gives a quadratic, solved with the quadratic formula. A constant term, like + 5, would not change the critical points, because it moves the whole graph up without changing any slopes.
The calculator lists both critical points in its steps and shows the smaller one as the main result. If a = 0, it solves the simpler 2bx + c = 0 instead.
How to find critical points step by step
- 1Differentiate f(x). Use the power rule on each term.
- 2Set f′(x) = 0. The flat points are where the slope is zero.
- 3Solve for x. For a cubic this is a quadratic, so there can be 2, 1 or 0 real answers.
- 4Classify each point. Find f″(x). If f″ < 0 it is a maximum; if f″ > 0 it is a minimum; if f″ = 0 the test can’t decide.
Now try it yourself. Enter a, b and c in the calculator below to see both critical points.
Try it yourself
Pre-filled with the example — change any value.
Common mistakes to avoid
- Assuming every critical point is a max or min. Check with the second derivative or the signs of f′ on each side.
x³has a flat point at 0 that is neither. - Missing the endpoints. On a closed interval, the biggest or smallest value can also be at an end.
- Expecting real answers every time. If the discriminant
(2b)² − 4(3a)(c)is negative, there are no real critical points, and the calculator says so. - Reading only the main result. The calculator shows the smaller critical point as its answer. Look in the steps for the other one.
Tips and tricks
- Sign chart: pick a test x on each side of a critical point. If f′ goes from + to −, it is a maximum; from − to +, a minimum.
- Second derivatives: get them quickly with the higher-order derivative calculator.
- Plug back in. The critical point is an x-value. To get the height, put it back into f(x).
Where you’ll use it in real life
- Business: the number of items that gives the most profit or the lowest cost per item.
- Packaging: the biggest box you can fold from a flat sheet of card.
- Physics: the highest point of a ball’s flight, where its vertical speed is zero.
- Engineering: the point of greatest bending in a beam.
Quick summary
To find critical points, differentiate, set f′(x) = 0 and solve. Then use f″(x) to decide which points are maximums and which are minimums. Use the calculator above for any cubic ax³ + bx² + cx; both critical points are listed in its steps.
Worked example
A company models its weekly profit (in thousands of US $) as f(x) = x³ − 6x² + 9x, where x is thousands of units made. Where does the profit curve level off? Here a = 1, b = −6 and c = 9. The steps show both critical points, x = 1 (a local maximum, profit 4) and x = 3 (a local minimum, profit 0).
- 01f(x) = 1x³ + -6x² + 9x
- 02f'(x) = 3x² + -12x + 9 = 0
- 03Discriminant = -12² − 4(3)(9) = 36
- 04x = (−-12 ± √36) / 6
- 05x₁ = 3, x₂ = 1
- 06Critical points at x = 3 and x = 1
These numbers are pre-filled in the calculator above.
Frequently asked questions
How do I find critical points?
Take the derivative, set it equal to zero and solve for x. Also include any x where the derivative is undefined.
How do I know if a critical point is a maximum or a minimum?
Use the second derivative. If f″(x) is negative, the point is a local maximum; if positive, a local minimum; if zero, the test cannot decide.
Can a function have no critical points?
Yes. For example, f(x) = x³ + x has f′(x) = 3x² + 1, which is never zero, so its graph is never flat.
