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Henry Maltby, Patrick Corn, and Jimin Khim contributed

This function has critical points at \(x = 1\) and \(x = 3\)

A critical point of a continuous function \(f\) is a point at which the derivative is zero or undefined. Critical points are the points on the graph where the function's rate of change is altered—either a change from increasing to decreasing, in concavity, or in some unpredictable fashion. Critical points are useful for determining extrema and solving optimization problems.

Contents

• Definition
• Types of Critical Points
• See Also

A continuous function \(f\) with \(x\) in its domain has a critical point at that point \(x\) if it satisfies either of the following conditions:

1. \(f'(x) = 0;\)
2. \(f'(x)\) is undefined.

A critical point of a differentiable function \(f\) is a point at which the derivative is 0.

Find all critical points of \(f(x) = x^4 - 4x^3 + 16x\).

The derivative of \(f\) is

\[f'(x) = 4x^3 - 12x^2 + 16 = 4(x + 1)(x - 2)^2,\]

so the derivative is zero at \(x = -1\) and \(x = 2\). Since \(f'\) is defined on all real numbers, the only critical points of the function are \(x = -1\) and \(x = 2.\ _\square\)

A local extremum is a maximum or minimum of the function in some interval of \(x\)-values. An inflection point is a point on the function where the concavity changes (the sign of the second derivative changes). While any point that is a local minimum or maximum must be a critical point, a point may be an inflection point and not a critical point.

1. A critical point is a local maximum if the function changes from increasing to decreasing at that point and is a local minimum if the function changes from decreasing to increasing at that point.
2. A critical point is an inflection point if the function changes concavity at that point.
3. A critical point may be neither. This could signify a vertical tangent or a "jag" in the graph of the function.

The first derivative test provides a method for determining whether a point is a local minimum or maximum. If the function is twice-differentiable, the second derivative test could also help determine the nature of a critical point. However, if the second derivative has value \(0\) at the point, then the critical point could be either an extremum or an inflection point.

Classify the critical points of the following function:

\[f(x) = \begin{cases} 1 - (x+1)^2 & x < 0 \\ 2x & 0 \le x \le 1 \\ 3 - (x - 2)^2 & 1 < x \le 2 \\ 3 + (x - 2)^3 & x > 2. \end{cases}\]

The derivative of \(f\) is

\[f'(x) = \begin{cases} -2(x+1) & x < 0 \\ 2 & 0 \le x \le 1 \\ -2(x-2) & 1 < x \le 2 \\ 3(x - 2)^2 & x > 2. \end{cases}\]

Note that the derivative has value \(0\) at points \(x = -1\) and \(x = 2\). At \(x = 0\), the derivative is undefined, and therefore \(x = 0\) is a critical point. At \(x = 1\), the derivative is \(2\) when approaching from the left and \(2\) when approaching from the right, so since the derivative is defined \((\)and equal to \(2 \ne 0),\) \(x = 1\) is not a critical point.

The critical point \(x = -1\) is a local maximum. The critical point \(x = 0\) is a local minimum. The critical point \(x = 2\) is an inflection point. \(_\square\)

\(x = -1\) is a local maximum, and \(x = 2\) is a local minimum. \(x = -1\) is a local maximum, and \(x = 2\) is an inflection point. \(x = -1\) is a local minimum, and \(x = 2\) is an inflection point. \(x = -1\) is a local minimum, and \(x = 2\) is a local maximum. \(x = -1\) is an inflection point, and \(x = 2\) is a local maximum. Reveal the answer

Classify the critical points of \(f(x) = x^4 - 4x^3 + 16x\).

The correct answer is: \(x = -1\) is a local minimum, and \(x = 2\) is an inflection point.

• Inflection Points
• Local and Absolute Extrema
• Optimization Problems
• Second Derivative Test

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