Lecture 25: Local and Absolute Extrema of Functions

1.1 Extremum of the Function

1.1.1 Basic Concepts

#####Definition: Theextremumofthefunction

description: if $\exists\delta>0$, make it

1) $\forall x\in U(x_0,\delta)$ always has $f(x)\geq f(x_0)$, then $f(x)$ is said to take the minimum value at $x_{0}$.

2) $\forall x\in U(x_0, \delta)$ always has $f(x)\leq f(x_0)$, then it is said that $f(x)$ takes the maximum value in $x_0$

#####Theorem: Extremaisanecessarycondition

description: If $f(x)$ is differentiable at $x_0$ and extremum is obtained at $x_0$, then $f^{\prime}(x_0)=0$

Significance

• For differentiable functions, this theorem narrows down the range of possible extremum points of the function;
• 1. For differentiable functions, only the stationary point needs to be considered;
• 2. For non-differentiable points, just look at stationary points and non-existent points;

The Relationship Between Extremals and Stations

• Location:
• The point of the derivative equal to zero is called thestationary point of the function;
• Take the extreme value at $x_0$, whose derivative is zero;
• The relationship between stationary points and extreme points
• 
• For differentiable functions, the extremum is the stationary point;

Corollary 1: Possible extrema points of a general function

• 1. Points where the derivative equals 0;
• 2. Points where the derivative does not exist;

#####Theorem: Thefirstsufficientconditionforextremevalues

Description: Let $f$ be continuous at $x_0$ and differentiable in a punctured neighborhood of $x_0$.

1. If $f'(x)\geq0$ for $x<x_0$ and $f'(x)\leq0$ for $x>x_0$, then $f$ has a local maximum at $x_0$.

2. If $f'(x)\leq0$ for $x<x_0$ and $f'(x)\geq0$ for $x>x_0$, then $f$ has a local minimum at $x_0$.

3. If $f'$ has the same sign on both sides of $x_0$, then $f$ has no local extremum there.

Significance

• Determine whether a point is truly an extreme point;
• How to use:
• From the necessary condition (theorem) of extremum -> several possible extremum points -> the first sufficient condition (theorem) for extremums -> the true extremum point;

Explanation

• Geometry
• Cases 1 and 2 respectively:
• Advantages
• It can analyze points where the derivative equals 0, and points where the derivative does not exist;

#####Theorem: Thesecondsufficientconditionfortheextremum

description: $\text{ Let }\quad f^{\prime}(x_0)=0,f^{\prime\prime}(x_0)\neq0$

1) When $f^{\prime\prime}(x_0)<0$, $f(x)$ take the maximum value at $x_{0}$

2) When $f^{\prime\prime}(x_0)>0,\quad f(x)$ takes the minimum value at $x_{0}$

Explanation

• Meaning:
• Determine whether a point has reached an extreme value by using the value of thesecond derivativeof a point whose derivative equals 0;
• Limitations:
• Unable to analyze points where the derivative does not exist;

1.1.2 Example Problems

Example Question: $\text{ Find the inverse of }f(x)=x^3-3x^2-9x+5\text{ Extremes }$

• Analysis
• From the form, thefunction is a differentiable function -> The extremum of the differentiable function can only be taken at the stationary position of;
• But a stationary point is not necessarily an extreme point; You need to use sufficient conditions to determine whether a point with zero first derivative of -> changes its sign (positive or negative);
• Analysis
• Using the first sufficient condition
• $f(x)=3x^{2}-6x-9=3\left(x_{-2}^{2}x-3\right)=3(x-3)(x+1)=0\Longrightarrow x_{1}=-1,x_{2}=3$
• At $x=-1$, the function derivative ranges from positive to >, so it is the maximum point;
• At $x=3$, the function derivative ranges from negative to >, so it is the minimum value;
• Use the second sufficient condition
• $\begin{aligned}f^{\prime\prime}(x)=6x-6,\quad f^{\prime\prime}(-1)=-12<0,\quad \text{ Extremely large }\\f^{\prime\prime}(3)=12>0\quad \text{ Very small }\end{aligned}$
• Question Type: Theextremumofthefunction

1.2 Maximum and Minimum Values

#####Definition: Themaximumandminimumvaluesofthecontinuousfunction

Description: Find the absolute extremum of continuous function $f(x)$ on $[a, b]$

0) Establishing the objective function (for word problems)

1) Find the station, non-differentiable, and endpoint $x_1,x_2,x_3 ...$ of $f(x)$ within $(a,b)$;

2) Find the function values for each point

3) Compare the calculated value with the endpoint; the larger value is the maximum, the smaller is the minimum

Explanation

• If the continuous function f(x) has only a unique extremum point in (a,b) -> If it is maximal, it is a maximum; If it is small, it is a minimum;

Supplement: Word problems for maximum-minimum values

• Step 1: Create the objective function $y=f(x)$
• Convert into general questions

Example: $\text{Find the maximum and minimum values of }f(x)=2x^3-3x^2\text{ on }[-1,2].$

• Analysis
• Since it is a polynomial, there are no points where the derivative does not exist, so only points where the derivative equals zero are needed;
• Analysis
• $f^{\prime}(x)=6x^2-6x=6x(x-1)=0\Longrightarrow x_1=0,x_2=1$
• Find the value of the stationary point
• $f(0)=0,f(1)=-1$
• Find the value of the endpoint
• $f(-1)=-5,f(2)=4$
• Therefore, x = -1 is the minimum point, and x = 2 is the maximum point
• Question Type: Themaximumandminimumvaluesofthecontinuousfunction

1.3 Related Question Types

Question Type: Findtheextremumandextremaofthefunction

PART 1: Problem-solving methods

Core: Composed of the necessary conditions (theorem) for extremums -> several possible extremum points, -> first sufficient condition (theorem) -> true extremum points;

Possible Extremes

• 1. Points where the derivative equals 0;
• 2. Points where the derivative does not exist;

Determining whether a possible extreme point is truly an extreme point

• When the derivative exists: the first sufficient condition of the extreme value can be used. -> Whether the left and right limits of the derivative change in signs. -> When the derivative changes from positive to negative: maximum + from negative to positive: minimum;
• The derivative does not exist:
• 1. When the function is continuous: you can use the first sufficient condition for the extreme value + continuous derivative -> You can derive whether the point where the derivative does not exist is an extremum -> When the derivative changes from positive to negative: maximum + from negative to positive: minimum value;
• 2. Unclear whether the function is continuous: use the first sufficient condition of the extremum + determine if this point is continuous (whether the left and right limits are equal) -> If continuous -> derivative changes from positive to negative: maximum + derivative from negative to positive: minimum;
• Or use the second sufficient condition for the extremum -> The second derivative is not zero;

Supplement: About the derivatives of functions and the left-right derivatives

• If one of the left and right derivatives of the derivative does not exist, then this derivative does not exist;
• And if one of the left and right derivatives has already been determined, then the other half does not need to be discussed;

Finding Derivatives Based on Function Formulas

• 1. Analyze the function and see if there are points where the derivative does not exist. -> Usually, the boundary points and zero points around the piecewise function may be points where the derivative does not exist;
• 2. After removing the nonexistent points, derivative is taken from the function;

Application Problems for Maximum/Minimum: First, establish a goal, then solve it using methods to find maximum/minimum.

• Note: Objective functions are usually not unique; consider whether there are simpler objective functions;

PART 2: Typical Example Problems

Example: Given that $f(x)$ is continuous in a neighborhood of $x=0$ and $f(0)=0,\lim_{x\to0}\frac{f(x)}{1-\cos x}=2$, then at point $x=0$ $f(x)$ __

• Analysis
• Analysis
• Direct Method:
• Since $\lim_{x\to0}\frac{f(x)}{1-\cos x}=2$ and $1-\cos x$ is greater than 0 -> near $x=0$, so near 0, $f(x)>0$;
• Because at 0 o'clock, $f(0)=0$, and near 0 again, $f(x)>0$;
• Therefore, it can be known that $f(0)=0$ is the smallest point near point 0, so the minimum value is obtained;
• Elimination method:
• substitute specific functions and analyze;
• Let $f(x)=x^2$, at which point it meets all the conditions of the question;
• From this, $f(x)=x^2$ finds its minimum value at point 0, and the limit exists, is differentiable, and its derivative equals 0;
• Question Type: #

PART 3: Key Points Review