This is a beta course, so its structure, chapters, and examples may continue to change.

Determination of Extrema

Graphical Method

Basic Idea

The most intuitive way to locate extrema is to look at the graph:

Maximum point: the graph reaches a local highest point there
Minimum point: the graph reaches a local lowest point there

When to Use It

The graph is easy to draw
You need a quick estimate of where extrema occur
As a verification for other methods

Caveats

A graph gives intuition, not a rigorous proof
For complicated functions the picture may be imprecise
Combine it with other methods to confirm

Numerical Comparison Method

Basic Idea

For a given point, compare nearby function values:

Maximum: if $f(x_0)$ is greater than all other values in its neighborhood
Minimum: if $f(x_0)$ is less than all other values in its neighborhood

Steps

Choose test points: pick a few points near the suspected extremum
Evaluate: compute the function values at those points
Compare: compare them with the value at the suspected point
Conclude: decide whether it is an extremum

Example

For $f(x) = x^2 - 4x + 3$ , suspect $x = 2$ :

$f(2) = -1$
$f(1) = 0 > -1$
$f(3) = 0 > -1$

So $x = 2$ is a minimum point.

The numerical comparison method is simple and intuitive, but you must choose suitable test points, and it cannot guarantee finding all extrema.

Derivative Method (Advanced)

The derivative method belongs to differential calculus and applies when the function is differentiable. It is the most rigorous method.

Necessary Condition

If $f$ is differentiable at $x_0$ and attains an extremum there, then $f'(x_0) = 0$ .

Sufficient Conditions

First Sufficient Condition

Let $f(x)$ be continuous on a neighborhood of $x_0$ and differentiable on a punctured neighborhood:

Maximum: $f'(x) > 0$ for $x < x_0$ , and $f'(x) < 0$ for $x > x_0$
Minimum: $f'(x) < 0$ for $x < x_0$ , and $f'(x) > 0$ for $x > x_0$

Second Sufficient Condition

Let $f(x)$ be twice differentiable at $x_0$ with $f'(x_0) = 0$ :

Maximum: $f''(x_0) < 0$
Minimum: $f''(x_0) > 0$

Examples

Example 1

Find the extrema of the function $f(x) = x^3 - 3x^2 + 2$ .

Reference Answer(5 个标签)

extremaderivative methodstationary pointsmaximumminimum

Solution Approach: Use the derivative method to find extrema, combined with graphical method for verification.

Detailed Steps:

Find the derivative: $f'(x) = 3x^2 - 6x = 3x(x - 2)$
Find stationary points: Set $f'(x) = 0$ , get $x = 0$ or $x = 2$
Determine extrema type:
- When $x < 0$ , $f'(x) > 0$ (function increasing)
- When $0 < x < 2$ , $f'(x) < 0$ (function decreasing)
- When $x > 2$ , $f'(x) > 0$ (function increasing)
Conclusion:
- $x = 0$ is a maximum point
- $x = 2$ is a minimum point
Calculate extrema values:
- $f(0) = 2$ (maximum)
- $f(2) = 8 - 12 + 2 = -2$ (minimum)

Answer: The function has a maximum value of $2$ at $x = 0$ and a minimum value of $-2$ at $x = 2$ .

Example 2

Find the extrema of the function $f(x) = |x|$ .

Reference Answer(4 个标签)

extremanon-differentiable functionnumerical comparison methodminimum

Solution Approach: This is a non-differentiable function that requires special handling.

Detailed Steps:

Analyze function properties: $f(x) = |x|$ is non-differentiable at $x = 0$ but continuous.
Use numerical comparison method:
- $f(0) = 0$
- For any $x \neq 0$ , $f(x) > 0$
Conclusion: $x = 0$ is a minimum point with minimum value $0$ .
Verification: The function reaches its global minimum at $x = 0$ .

Answer: The function has a minimum value of $0$ at $x = 0$ .

Exercises

Exercise 1

Find the extrema of the function $f(x) = x^4 - 4x^2$ .

Reference Answer(5 个标签)

extremaderivative methodmultiple stationary pointsmaximumminimum

Solution Approach: Use the derivative method to find extrema.

Detailed Steps:

Find the derivative: $f'(x) = 4x^3 - 8x = 4x(x^2 - 2)$
Find stationary points: Set $f'(x) = 0$ , get $x = 0$ or $x = \pm\sqrt{2}$
Determine extrema type:
- When $x < -\sqrt{2}$ , $f'(x) < 0$
- When $-\sqrt{2} < x < 0$ , $f'(x) > 0$
- When $0 < x < \sqrt{2}$ , $f'(x) < 0$
- When $x > \sqrt{2}$ , $f'(x) > 0$
Conclusion:
- $x = -\sqrt{2}$ is a minimum point
- $x = 0$ is a maximum point
- $x = \sqrt{2}$ is a minimum point
Calculate extrema values:
- $f(-\sqrt{2}) = 4 - 8 = -4$ (minimum)
- $f(0) = 0$ (maximum)
- $f(\sqrt{2}) = 4 - 8 = -4$ (minimum)

Answer: The function has a maximum value of $0$ at $x = 0$ and minimum values of $-4$ at $x = \pm\sqrt{2}$ .

Exercise 2

Find the extrema of the function $f(x) = \sin x + \cos x$ on the interval $[0, 2\pi]$ .

Reference Answer(5 个标签)

extrematrigonometric functionsinterval extremamaximumminimum

Solution Approach: Use the derivative method to find extrema.

Detailed Steps:

Find the derivative: $f'(x) = \cos x - \sin x$
Find stationary points: Set $f'(x) = 0$ , get $\cos x = \sin x$ That is $\tan x = 1$ , so $x = \frac{\pi}{4} + k\pi$ , $k \in \mathbb{Z}$ On $[0, 2\pi]$ , stationary points are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$
Determine extrema type:
- When $0 < x < \frac{\pi}{4}$ , $f'(x) > 0$
- When $\frac{\pi}{4} < x < \frac{5\pi}{4}$ , $f'(x) < 0$
- When $\frac{5\pi}{4} < x < 2\pi$ , $f'(x) > 0$
Conclusion:
- $x = \frac{\pi}{4}$ is a maximum point
- $x = \frac{5\pi}{4}$ is a minimum point
Calculate extrema values:
- $f(\frac{\pi}{4}) = \sin\frac{\pi}{4} + \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$ (maximum)
- $f(\frac{5\pi}{4}) = \sin\frac{5\pi}{4} + \cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}$ (minimum)

Answer: The function has a maximum value of $\sqrt{2}$ at $x = \frac{\pi}{4}$ and a minimum value of $-\sqrt{2}$ at $x = \frac{5\pi}{4}$ .

Extrema of Functions

Local maximum and minimum values, how to locate extremum points by graphing and numerical comparison, with worked examples.

Convexity and Concavity of Functions

Concave and convex functions, second-derivative tests, and applications such as Jensen's inequality, with worked examples.