"Exercise Course Application of Basic Inequalities" Quadratic Functions, Equations and Inequalities of One Variable PPT

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"Exercise Course Application of Basic Inequalities" Quadratic Functions, Equations and Inequalities of One Variable PPT

Part One: Explanation of Curriculum Standards

1. Be able to use basic inequalities to find the optimal value of a function and the optimal value of an algebraic expression.

2. Be able to use basic inequalities to solve optimal value problems in practical problems.

PPT on the application of basic inequalities in exercise lessons, part 2: independent preview

Use basic inequalities to find the optimal value of functions, algebraic expressions, and practical problems

1. (1) What are the conditions for the application of the basic inequality √ab≤(a+b)/2?

Tip: One is positive, two is definite, and three is equal, that is: ① a and b are both positive numbers; ② one of a+b and ab is a definite value; ③ the equal sign in the inequality must be obtained.

(2) It is known that (a+b)/2≥√ab, where a>0, b>0, if ab is a constant P, how does the value of a+b change?

(3) If a+b is constant S, how does the value of ab change?

Tip: If and only if a=b, ab has a maximum value of 1/4S2.

2. Fill in the blanks

Equivalent deformation of the formula: ab≤(a^2+b^2)/2,ab≤ (a+b)/2 2.

3. Do it

(1) The maximum value of function f(x)=x+ (x<0) is ________;

(2) If the positive numbers a and b satisfy 2a+3b=8, then the maximum value of ab is ________.

Analysis: (1) Since x<0, so f(x)=x+9/x=-["(-" x")" +("-" 9/x)]≤- 2√("(-" x")•" ("-" 9/x) )=-6, if and only if -x=-9/x, that is, x=-3, the function takes the maximum value -6 .

(2) Since a, b>0, so ab=1/6•2a•3b≤1/6•((2a+3b)/2)^2=8/3, if and only if 2a=3b, that is When a=2,b=4/3, ab takes the maximum value 8/3.

PPT on the application of basic inequalities in exercise lessons, part three: inquiry learning

Study 1: Use basic inequalities to find the optimal value of functions and algebraic expressions

1. Find the optimal value by applying basic inequalities after deformation

Example 1 Find the maximum value of the following function and find the corresponding x value.

(1)y=x+1/2x(x<0);

(2)y=1/(x"-" 3)+x(x>3);

(3)y=x(1-3x) 0

Solution: (1)y=x+1/2x=- (-x)+1/(2"(-" x")" ) ≤-2√("(-" x")•" 1/(2 "(-" x")" ))=-√2, if and only if x=1/2x(x<0), that is, when x=-√2/2, y takes the maximum value -√2.

(2)y=1/(x"-" 3)+x=1/(x"-" 3)+(x-3)+3≥2√(1/(x"-" 3) "•( " x"-" 3")" )+3=5, if and only if 1/(x"-" 3)=x-3(x>3), that is, when x=4, y takes the minimum value 5.

(3)y=x(1-3x)=1/3×3x(1-3x)≤1/3× (3x+"(" 1"-" 3x")" )/2 2=1/12, when And only when 3x=1-3x, that is, x=1/6, y takes the maximum value 1/12.

Reflect and understand that the key to using basic inequalities to find the optimal value is to obtain the fixed value conditions. When solving problems, you should refer to the known conditions and desired formulas, and use appropriate methods such as "splitting terms, adding terms, matching, and deformation" to create basic values. The conditions of the inequality can be summarized as follows: one is not positive, use its opposite to change the direction of the inequality sign; two is indefinite, the definite sum or product should be made up; three is not equal, generally other methods need to be used, such as trying to use the monotonicity of the function (Learn in Chapter 3).

PPT on the application of basic inequalities in exercise lessons, part 4: in-class exercises

1. The maximum value of the function y=2x(2-x) (where 0

A.1/4 B.1/2 C.1 D.2

Analysis: ∵0

Answer:D

2. Assume x>0, y>0, x+y=4, then the minimum value of 1/x+4/y is _______.

Analysis:∵x+y=4,∴1/x+4/y=1/4 1/x+4/y (x+y)=1/4 5+y/x+4x/y,

And x>0, y>0, then y/x+4x/y≥2√(y/x "•" 4x/y)=4 Take the equal sign if and only if y/x=4x/y,

Then 1/x+4/y≥1/4×(5+4)=9/4.

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Update Time: 2024-10-17

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