"End of Chapter Review Lesson" Quadratic functions, equations and inequalities of one variable PPT
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"End of Chapter Review Lesson" Quadratic functions, equations and inequalities of one variable PPT

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"End of Chapter Review Lesson" Quadratic functions, equations and inequalities of one variable PPT

reminder to explore

Properties of Inequalities

[Example 1] If a, b, c satisfy c<b<a and ac<0, then one of the following options may not be true ()

A. ab>acB. c(b-a)>0

C. cb2<ab2 D. ac(a-c)<0

C c<b<a, ac<0⇒a>0, c<0.

For A: b>ca>0⇒ab>ac, A is correct.

For B: b<a⇒b-a<0c<0⇒c·(b-a)>0, B is correct.

For C: c<ab2≥0⇒cb2≤ab2 cb2<ab2, C is wrong, that is, C is not necessarily true.

For D: ac<0, a-c>0⇒ac(a-c)<0, D is correct, choose C.]

regular method

The true or false judgment of an inequality depends on its scope of application and conditions. Giving counterexamples is a good way to judge whether a proposition is false. When verifying with the special case method, you should pay attention to that what is suitable is not necessarily right, and what is not suitable is definitely wrong, so special cases You can only negate the choices. As long as three out of the four are eliminated, the remaining ones are the correct answers.

Track training

1. If a>b>c and a+b+c=0, then the correct one of the following inequalities is ()

A. ab>ac

B. ac>bc

C. a|b|>c|b|

D. a2>b2>c2

2. If 1≤a≤5, -1≤b≤2, then the value range of a-b is ________.

basic inequalities

[Example 2] Assume x<-1, find the maximum value of y=x+5x+2x+1.

[Solution] ∵x<-1, ∴x+1<0.

∴-(x+1)>0,

∴y=x+5x+2x+1=x2+7x+10x+1

=x+12+5x+1 +4x+1=(x+1)+4x+1+5

=--x+1+4-x+1+5

≤-24+5=1,

When (x+1)2=4, that is, x=-3, take "=". ]

regular method

The main application of basic inequalities is to find the maximum value or range of a function, which is applicable to both the case of one variable and the case of two variables. Basic inequalities have the functions of converting "sum expression" into "product expression" and converting "product expression" ” is transformed into the scaling function of “sum”. The key to solving such problems is to create conditions for applying inequalities. Reasonably splitting terms or matching factors are common problem-solving techniques, and the purpose of splitting and combining is to make the equal sign can be established.

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