Critical Point
In short: Critical point names an input that belongs to a function’s domain and makes the slope either flat or undefined, so a peak or valley could be hiding there.
AP Calculus ABCalculusAP Calculus AB Unit 5
What is Critical Point?
A critical point marks a spot where the graph might turn around. At such an input the tangent line is flat, or the curve has a sharp corner or vertical tangent so that no ordinary tangent slope exists.
Formally, \(x = c\) is a critical number of \(f\) when \(c\) is in the domain of \(f\) and either \(f^{\prime}(c) = 0\) or \(f^{\prime}(c)\) is undefined. Being critical is only a necessary sign of a peak or a valley: some critical points, like the flat spot of \(x^3\) at 0, are neither.
How to spot it
- A question asks where a graph might top out or bottom out locally.
- The instructions say to locate every input where the slope vanishes or breaks down.
- A graph shows a flat tangent, a sharp corner (cusp) or a vertical tangent.
- Optimization or increasing-decreasing analysis is the goal, so you need the list of dividing inputs first.
The method
- Find the domain of \(f\) and compute \(f^{\prime}(x)\).
- Solve \(f^{\prime}(x) = 0\) to get the flat-tangent inputs.
- Find the inputs where \(f^{\prime}(x)\) is undefined (zero denominators, even roots of negative values) but \(f\) itself is defined.
- List every such \(x\) as a critical point, then test it with the first or second derivative test.
Critical Point: solved example
Problem
Find the critical points of \(f(x) = 2x^3 – 9x^2 – 24x + 7\).
Solution
The function is a polynomial, so its domain is all real numbers and the derivative exists everywhere.
\(f^{\prime}(x) = 6x^2 – 18x – 24 = 6(x^2 – 3x – 4) = 6(x – 4)(x + 1)\).
Setting it equal to zero gives \(x = -1\) and \(x = 4\). Then \(f(-1) = -2 – 9 + 24 + 7 = 20\) and \(f(4) = 128 – 144 – 96 + 7 = -105\).
Try it
Question 1
Find the critical number of \(g(x) = (x – 2)^{2/3}\).
Show the answer
\(g^{\prime}(x) = \frac{2}{3}(x – 2)^{-1/3}\) is never zero, but it is undefined at \(x = 2\), where \(g(2) = 0\) exists. So \(x = 2\) is the only critical number.
Question 2
Is \(x = 0\) a critical number of \(f(x) = \frac{1}{x}\)?
Show the answer
No. Although \(f^{\prime}\) is undefined at 0, the function itself is not defined there, so 0 is not in the domain.
Critical Point vs. Inflection Point
A critical point is found with the first derivative and concerns peaks, valleys and flat spots. An inflection point is found with the second derivative and marks where the bending direction changes. A graph can have one without the other.
Keep going
Lessons
Worksheets and quizzes
- Increasing, Decreasing and the First Derivative Test – Worksheet
- Increasing, Decreasing and the First Derivative Test – Quiz
Related words
- First Derivative Test
- Second Derivative Test
- Extreme Value Theorem
- Optimization (Calculus)
- Derivative
Whole dictionary: A–Z · AP Calculus AB · Calculus