# Critical Points

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.

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## 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.

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 alocal minimum.

The critical point $x = 2$ is aninflection point. $_\square$

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