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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

extremaderivative methodmaximumminimum

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

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

extremanon-differentiable functionnumerical comparison methodminimum

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

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

extremaderivative methodmaximumminimum

Approach: Use the derivative method to find extrema.

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

extrematrigonometric functionmaximumminimum

Approach: Use the derivative method to find extrema.

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 $x = \frac{5\pi}{4}$ .

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$f(x)$	Mathematical symbol	f of x	The function whose extrema are being located
$f'(x)$	Mathematical symbol	f prime of x	The first derivative whose sign on each side of $x_0$ decides maximum or minimum
$f''(x)$	Mathematical symbol	f double prime of x	The second derivative used in the second sufficient condition
$x_0$	Mathematical symbol	x zero	The candidate extremum point where the condition $f'(x_0) = 0$ is checked
$f(x_0)$	Mathematical symbol	f of x zero	The value compared with nearby values in the numerical comparison method
$\lvert x \rvert$	Mathematical symbol	absolute value of x	The non-differentiable example whose minimum value $0$ sits at $x = 0$
$\pm\sqrt{2}$	Mathematical symbol	plus or minus root two	The two minimum points found in Exercise 1, where the value is $-4$
$\frac{\pi}{4}$	Mathematical symbol	pi over four	The maximum point of $\sin x + \cos x$ on $[0, 2\pi]$
$k \in \mathbb{Z}$	Mathematical symbol	k in the integers	The integer parameter in the general solution $x = \frac{\pi}{4} + k\pi$
$[0, 2\pi]$	interval	closed interval from zero to two pi	The closed interval on which the extrema of $\sin x + \cos x$ are searched
$\tan x$	Mathematical symbol	tangent of x	The equation $\tan x = 1$ that locates the stationary points in Exercise 2
$\sin x + \cos x$	Mathematical symbol	sine x plus cosine x	The function whose extrema on $[0, 2\pi]$ are $\sqrt{2}$ and $-\sqrt{2}$

Extrema of Functions

Convexity and Concavity of Functions

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Determination of Extrema

Graphical Method

Basic Idea

When to Use It

Caveats

Numerical Comparison Method

Basic Idea

Steps

Example

Derivative Method (Advanced)

Necessary Condition

Sufficient Conditions

First Sufficient Condition

Second Sufficient Condition

Examples

Example 1

Example 2

Exercises

Exercise 1

Exercise 2

Summary

Symbols Used in This Article

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