(共16张PPT)
函数零点与方程的解(1)
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“ Work Report ”
授课对象:高一(12)班
授课人:邓芝
CONTENTS
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学习目标
结合二次函数的图象,了解函数的零点与方程根的联系.
会求简单函数的零点、零点个数及零点所在的大致区间.
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思考
x=1
x=3
超越方程
有几个实数解,解的大致范围是什么?
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新知探索
一元二次方程 x2-2x-3=0 x2-2x+1=0 x2-2x+3=0
二次函数 y= x2-2x-3 y= x2-2x+1 y= x2-2x+3
函 数 的 图 像
方程的根 x1=-1,x2=3 x1=x2=1 无实数根
图像与x轴交点的横坐标 -1,3 1 无交点
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思考:方程的根与函数的图像和x轴交点的横坐标有什么关系?
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概念生成
函数y=f(x)的零点
函数的零点:把使方程f(x)=0成立的实数x叫做函数 y=f(x)的零点
注意:零点不是点,是一个数值
方程f(x)=0的实数根
1.若y=f(x)表示函数y=f(x)的图像与x轴有公共点
2.若f(x)=g(x)-h(x),表示y=h(x)与y=g(x)的公共点
代数意义
几何意义
1.所有函数都存在零点吗?
2.什么条件下零点存在,存在一定唯一吗?
画出二次函数y=x2+4x-5,观察图像,它的两个零点所在的大致区间是什么?
x
y
自主探究
1 (0,2)
3 (2,4)
由图可知,函数在区间(0,2)和区间(2,4)上有零点。在零点所在区间内,函数图像与x轴有什么关系?所在区间端点值有什么规律?
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情景探索
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实验探索
小实验
函数y=f(x)在区间(a,b)内为连续函数,且满足f(a)f(b)<0
·(a,f(a))
·(a,f(a))
·(a,f(a))
·(b,f(b))
·(b,f(b))
·(b,f(b))
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概念深化
函数零点存在性定理
如果函数y=f(x)在区间[a,b]上的图像是一条连续不断的曲线,且有f(a)f(b)<0,那么,函数y=f(x)在区间(a,b)内至少有一个零点,即存在c (a,b),使得f(c)=0,这个c也就是方程f(x)=0的解
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辨析比较
判断下列说法是否正确,如果不正确请举出反例
1.如果函数y=f(x)在区间[a,b]上的图像是一条连续不断的曲线,且在区间(a,b)内有零点,则可以推出来f(a)f(b)<0。
2.如果函数y=f(x)在区间[a,b]上满足f(a)f(b)<0,则f(x)在区间(a,b)上一定有零点。
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新旧结合
在你学过的所有函数中,哪个函数f(x)在区间R内恰有0个、1个、2个,......,甚至无数个零点?
试举例说明.
例1.求方程lnx+2x-6=0实数解的个数,解的大致范围是什么
解:
设函数f(x)=lnx+2x-6,则 f(x)的定义域为(0,+∞)
列表,并作出f(x)的图象
x 1 2 3 4 5 6 7 8 9
f(x) -4 -1.3069 1.0986 3.3863 5.6094 7.7918 9.9459 12.0794 14.1972
f(2)<0, f(3)>0,
即 f(2) f(3)<0.
由函数零点存在定理,
f(x)在(2,3)内至少有一个零点.
又∵ f(x)=lnx+2x-6是增函数,
∴ f(x)是只有一个零点,
方程lnx+2x-6=0有1个实数解.
由表和图象得
f(x)=lnx+2x-6
典例练习
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典例练习
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典例练习
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解:
典例练习
感谢の观看
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