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初三數(shù)學(xué)函數(shù)對(duì)稱性的探究課件

時(shí)間:2021-06-11 15:31:30 課件 我要投稿

初三數(shù)學(xué)函數(shù)對(duì)稱性的探究課件

  一、 函數(shù)自身的對(duì)稱性探究

初三數(shù)學(xué)函數(shù)對(duì)稱性的探究課件

  定理1.函數(shù) y = f (x)的圖像關(guān)于點(diǎn)A (a ,b)對(duì)稱的充要條件是

  f (x) + f (2a-x) = 2b

  證明:(必要性)設(shè)點(diǎn)P(x ,y)是y = f (x)圖像上任一點(diǎn),∵點(diǎn)P( x ,y)關(guān)于點(diǎn)A (a ,b)的對(duì)稱點(diǎn)P(2a-x,2b-y)也在y = f (x)圖像上, 2b-y = f (2a-x)

  即y + f (2a-x)=2b故f (x) + f (2a-x) = 2b,必要性得證。

  (充分性)設(shè)點(diǎn)P(x0,y0)是y = f (x)圖像上任一點(diǎn),則y0 = f (x0)

  ∵ f (x) + f (2a-x) =2bf (x0) + f (2a-x0) =2b,即2b-y0 = f (2a-x0) 。

  故點(diǎn)P(2a-x0,2b-y0)也在y = f (x) 圖像上,而點(diǎn)P與點(diǎn)P關(guān)于點(diǎn)A (a ,b)對(duì)稱,充分性得征。

  推論:函數(shù) y = f (x)的圖像關(guān)于原點(diǎn)O對(duì)稱的充要條件是f (x) + f (-x) = 0

  定理2.函數(shù) y = f (x)的圖像關(guān)于直線x = a對(duì)稱的充要條件是

  f (a +x) = f (a-x) 即f (x) = f (2a-x) (證明留給讀者)

  推論:函數(shù) y = f (x)的圖像關(guān)于y軸對(duì)稱的充要條件是f (x) = f (-x)

  定理3. ①若函數(shù)y = f (x) 圖像同時(shí)關(guān)于點(diǎn)A (a ,c)和點(diǎn)B (b ,c)成中心對(duì)稱(ab),則y = f (x)是周期函數(shù),且2| a-b|是其一個(gè)周期。

 、谌艉瘮(shù)y = f (x) 圖像同時(shí)關(guān)于直線x = a 和直線x = b成軸對(duì)稱(ab),則y = f (x)是周期函數(shù),且2| a-b|是其一個(gè)周期。

 、廴艉瘮(shù)y = f (x)圖像既關(guān)于點(diǎn)A (a ,c) 成中心對(duì)稱又關(guān)于直線x =b成軸對(duì)稱(ab),則y = f (x)是周期函數(shù),且4| a-b|是其一個(gè)周期。

  ①②的證明留給讀者,以下給出③的證明:

  ∵函數(shù)y = f (x)圖像既關(guān)于點(diǎn)A (a ,c) 成中心對(duì)稱,

  f (x) + f (2a-x) =2c,用2b-x代x得:

  f (2b-x) + f [2a-(2b-x) ] =2c(*)

  又∵函數(shù)y = f (x)圖像直線x =b成軸對(duì)稱,

  f (2b-x) = f (x)代入(*)得:

  f (x) = 2c-f [2(a-b) + x](**),用2(a-b)-x代x得

  f [2 (a-b)+ x] = 2c-f [4(a-b) + x]代入(**)得:

  f (x) = f [4(a-b) + x],故y = f (x)是周期函數(shù),且4| a-b|是其一個(gè)周期。

  二、 不同函數(shù)對(duì)稱性的探究

  定理4.函數(shù)y = f (x)與y = 2b-f (2a-x)的圖像關(guān)于點(diǎn)A (a ,b)成中心對(duì)稱。

  定理5.①函數(shù)y = f (x)與y = f (2a-x)的圖像關(guān)于直線x = a成軸對(duì)稱。

 、诤瘮(shù)y = f (x)與a-x = f (a-y)的圖像關(guān)于直線x +y = a成軸對(duì)稱。

 、酆瘮(shù)y = f (x)與x-a = f (y + a)的圖像關(guān)于直線x-y = a成軸對(duì)稱。

  定理4與定理5中的①②證明留給讀者,現(xiàn)證定理5中的③

  設(shè)點(diǎn)P(x0 ,y0)是y = f (x)圖像上任一點(diǎn),則y0 = f (x0)。記點(diǎn)P( x ,y)關(guān)于直線x-y = a的軸對(duì)稱點(diǎn)為P(x1, y1),則x1 = a + y0 , y1 = x0-a ,x0 = a + y1 , y0= x1-a 代入y0 = f (x0)之中得x1-a = f (a + y1) 點(diǎn)P(x1, y1)在函數(shù)x-a = f (

  (y + a)的圖像上。

  同理可證:函數(shù)x-a = f (y + a)的圖像上任一點(diǎn)關(guān)于直線x-y = a的軸對(duì)稱點(diǎn)也在函數(shù)y = f (x)的圖像上。故定理5中的③成立。

  推論:函數(shù)y = f (x)的圖像與x = f (y)的.圖像關(guān)于直線x = y 成軸對(duì)稱。

  三、 三角函數(shù)圖像的對(duì)稱性列表

  函 數(shù)

  對(duì)稱中心坐標(biāo)

  對(duì)稱軸方程

  y = sin x

  ( k, 0 )

  x = k/2

  y = cos x

  ( k/2 ,0 )

  x = k

  y = tan x

  (k/2 ,0 )

  無

  注:①上表中kZ

 、趛 = tan x的所有對(duì)稱中心坐標(biāo)應(yīng)該是(k/2 ,0 ),而在岑申、王而冶主編的浙江教育出版社出版的21世紀(jì)高中數(shù)學(xué)精編第一冊(cè)(下)及陳兆鎮(zhèn)主編的廣西師大出版社出版的高一數(shù)學(xué)新教案(修訂版)中都認(rèn)為y = tan x的所有對(duì)稱中心坐標(biāo)是( k, 0 ),這明顯是錯(cuò)的。

  四、 函數(shù)對(duì)稱性應(yīng)用舉例

  例1:定義在R上的非常數(shù)函數(shù)滿足:f (10+x)為偶函數(shù),且f (5-x) = f (5+x),則f (x)一定是( )(第十二屆希望杯高二第二試題)

  (A)是偶函數(shù),也是周期函數(shù) (B)是偶函數(shù),但不是周期函數(shù)

  (C)是奇函數(shù),也是周期函數(shù) (D)是奇函數(shù),但不是周期函數(shù)

  解:∵f (10+x)為偶函數(shù),f (10+x) = f (10-x).

  f (x)有兩條對(duì)稱軸 x = 5與x =10 ,因此f (x)是以10為其一個(gè)周期的周期函數(shù),x =0即y軸也是f (x)的對(duì)稱軸,因此f (x)還是一個(gè)偶函數(shù)。

  故選(A)

  例2:設(shè)定義域?yàn)镽的函數(shù)y = f (x)、y = g(x)都有反函數(shù),并且f(x-1)和g-1(x-2)函數(shù)的圖像關(guān)于直線y = x對(duì)稱,若g(5) = 1999,那么f(4)=( )。

  (A) 1999; (B)2000; (C)2001;(D)2002。

  解:∵y = f(x-1)和y = g-1(x-2)函數(shù)的圖像關(guān)于直線y = x對(duì)稱,

  y = g-1(x-2) 反函數(shù)是y = f(x-1),而y = g-1(x-2)的反函數(shù)是:y = 2 + g(x), f(x-1) = 2 + g(x), 有f(5-1) = 2 + g(5)=2001

  故f(4) = 2001,應(yīng)選(C)

  例3.設(shè)f(x)是定義在R上的偶函數(shù),且f(1+x)= f(1-x),當(dāng)-10時(shí),

  f (x) = - x,則f (8.6 ) = _________ (第八屆希望杯高二第一試題)

  解:∵f(x)是定義在R上的偶函數(shù)x = 0是y = f(x)對(duì)稱軸;

  又∵f(1+x)= f(1-x) x = 1也是y = f (x) 對(duì)稱軸。故y = f(x)是以2為周期的周期函數(shù),f (8.6 ) = f (8+0.6 ) = f (0.6 ) = f (-0.6 ) = 0.3

  例4.函數(shù) y = sin (2x + )的圖像的一條對(duì)稱軸的方程是( )(92全國(guó)高考理) (A) x = - (B) x = - (C) x = (D) x =

  解:函數(shù) y = sin (2x + )的圖像的所有對(duì)稱軸的方程是2x + = k+

  x = - ,顯然取k = 1時(shí)的對(duì)稱軸方程是x = - 故選(A)

  例5. 設(shè)f(x)是定義在R上的奇函數(shù),且f(x+2)= -f(x),當(dāng)01時(shí),

  f (x) = x,則f (7.5 ) = ( )

  (A) 0.5 (B) -0.5 (C) 1.5 (D) -1.5

  解:∵y = f (x)是定義在R上的奇函數(shù),點(diǎn)(0,0)是其對(duì)稱中心;

  又∵f (x+2 )= -f (x) = f (-x),即f (1+ x) = f (1-x),直線x = 1是y = f (x) 對(duì)稱軸,故y = f (x)是周期為2的周期函數(shù)。

  f (7.5 )

  = f (8-0.5 ) = f (-0.5 ) = -f (0.5 ) =-0.5 故選(B)

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