Applications of Derivatives - Core Definitions

Critical points and extrema are central to optimization problems across mathematics, science, and engineering. Understanding where functions achieve their maximum and minimum values is essential for modeling real-world phenomena.

DefinitionCritical Points

A critical point of a function $f$ is a point $c$ in the domain of $f$ where either:

$f'(c) = 0$ , or
$f'(c)$ does not exist

Critical points are candidates for local extrema, though not every critical point corresponds to an extremum.

DefinitionLocal and Absolute Extrema

Let $f$ be defined on a domain $D$ .

$f$ has a local maximum at $c$ if $f(c) \geq f(x)$ for all $x$ in some open interval containing $c$
$f$ has a local minimum at $c$ if $f(c) \leq f(x)$ for all $x$ in some open interval containing $c$
$f$ has an absolute maximum at $c$ if $f(c) \geq f(x)$ for all $x \in D$
$f$ has an absolute minimum at $c$ if $f(c) \leq f(x)$ for all $x \in D$

Collectively, maxima and minima are called extrema.

TheoremFermat's Theorem

If $f$ has a local extremum at $c$ and $f'(c)$ exists, then $f'(c) = 0$ .

Remark

Fermat's Theorem provides a necessary condition for extrema: they can only occur at critical points. However, it is not sufficient—the function $f(x) = x^3$ has $f'(0) = 0$ , but $x = 0$ is neither a local maximum nor minimum.

ExampleFinding Critical Points

Find all critical points of $f(x) = x^3 - 6x^2 + 9x + 1$ .

First, compute the derivative: $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$

Setting $f'(x) = 0$ : $3(x - 1)(x - 3) = 0$

The critical points are $x = 1$ and $x = 3$ . Since $f$ is a polynomial, $f'$ exists everywhere, so these are the only critical points.

DefinitionFirst Derivative Test

Let $c$ be a critical point of a continuous function $f$ .

If $f'$ changes from positive to negative at $c$ , then $f$ has a local maximum at $c$
If $f'$ changes from negative to positive at $c$ , then $f$ has a local minimum at $c$
If $f'$ does not change sign at $c$ , then $f$ has neither a maximum nor minimum at $c$

ExampleApplying the First Derivative Test

Classify the critical points of $f(x) = x^3 - 6x^2 + 9x + 1$ .

We found critical points at $x = 1$ and $x = 3$ . Test the sign of $f'(x) = 3(x-1)(x-3)$ :

For $x < 1$ : choose $x = 0$ , so $f'(0) = 3(−1)(−3) = 9 > 0$ (increasing)
For $1 < x < 3$ : choose $x = 2$ , so $f'(2) = 3(1)(−1) = −3 < 0$ (decreasing)
For $x > 3$ : choose $x = 4$ , so $f'(4) = 3(3)(1) = 9 > 0$ (increasing)

Therefore:

At $x = 1$ : $f'$ changes from positive to negative, so $f$ has a local maximum
At $x = 3$ : $f'$ changes from negative to positive, so $f$ has a local minimum

Understanding critical points and extrema is fundamental to optimization, allowing us to find maximum efficiency, minimum cost, or optimal design parameters in countless applications.