Critical Points
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
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:
- \(f'(x) = 0;\)
- \(f'(x)\) 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 \(x = 2.\ _\square\)
Types of Critical Points
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.
- 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.
- A critical point is an inflection point if the function changes concavity at that point.
- 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.
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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. \(_\square\)
Classify the critical points of \(f(x) = x^4 - 4x^3 + 16x\).