§3.基本初等函数 1.把下列在[0,1)上定义的函数延拓到整个实轴上去,使它成为以1为周期的函数: (3)y=e2 (1)延拓后的函数为y=(x-n)2(n≤x<n+1,n∈Z) (2)延拓后的函数为y=sin(x-n)(n≤x<n+1,n∈Z) 3)延拓后的函数为y=e(n≤x<n+1,n∈Z) 2.把下列在[0,+∞)上定义的函数延拓到整个实轴上去,(a)使它们成为奇函数;(b)使它们成为偶函数 (1)y=x2 (2)y (1)延拓后的函数为 )f(x)=/x, x2,x<0 (b)f(x)=x2 (2)延拓后的函数为: (a)f(a)=sin:z b)f(a)=sin rl 3.做下列函数的图形 (1)y=sgn cos z 4.作函数y=(x)的图形 5.作函数y=]-r的图
11 §3. ƒ – ºÍ 1. re3[0, 1)˛½¬ºÍÚˇá¢¶˛ß¶ß§è±1豜ºÍµ (1) y = x 2 (2) y = sin x (3) y = e x )µ (1) ÚˇºÍèy = (x − n) 2 (n 6 x < n + 1, n ∈ Z) (2) ÚˇºÍèy = sin(x − n)(n 6 x < n + 1, n ∈ Z) (3) ÚˇºÍèy = e x−n (n 6 x < n + 1, n ∈ Z) 2. re3[0, +∞)˛½¬ºÍÚˇá¢¶˛ß(a)¶ßǧ褺Ͷ(b)¶ßǧèۺ͵ (1) y = x 2 (2) y = sin x )µ (1) ÚˇºÍèµ (a) f(x) = x 2 , x > 0 −x 2 , x < 0 (b) f(x) = x 2 (2) ÚˇºÍèµ (a) f(x) = sin x (b) f(x) = sin |x| 3. âeºÍ„/µ (1) y = sgn cos x (2) y = [x] − 2 h x 2 i )µ (1) ✲ ✻ π 0 π xy 1 -1 qq qq ❜❜ ❜ ❜ ❜ ❜ ❜ ❜ (2) ✲ ✻ 0 1 2 3-1-2-3 x y 1 ❝ ❜ ❜ ❜❜❜ ❜ 4. äºÍy = (x)„/. )µ ✲ ✻ 0 1 2 3-1-2-3 x y 1 ❝ ❜❜ ❜❜❜❜ 5. äºÍy = [x] − x„/. )µ ✻ ✲ ❅ ❅❅ ❅ ❅❅ ❅ ❅❅ ❅ ❅❅ ❅❅ ❅ ❅❅ ❅ ❅❅ ❅❅ 0 1 2 3-1-2-3 x y -1❝ ❜❜ ❜❜❜❜
6.一个函数是用下述方法决定的:在每一个小区间n≤x<n+1(其中n为整数)内f(x)是线性的且f(n)= 1,∫(n+5)=0,试作此函数的图形 7.作函数y=|sinx+2cosx|的图形. 8.若已知函数∫(x)=tanx,作下列函数的图形: (1)y=f(2x) (2)y=f(kx+b)(k≠0) (3)y=f (2)(k,b>0)
12 6. òáºÍ¥^e„ê{˚½µ3zòá´mn 6 x < n + 1(Ÿ•nèÍ)Sf(x)¥Ç5Öf(n) = −1, f n + 1 2 = 0ߣädºÍ„/. )µ ✲ ✻ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ 0 1 2 3-1-2-3 x y 1 -1 ❝ ❜❜ ❜❜❜❜ 7. äºÍy = | sin x + 2 cos x|„/. )µ ✲ ✻ 0 π 2 π − π 2 −π x y √ 5 -1 8. eƺÍf(x) = tan xßäeºÍ„/µ (1) y = f(2x) (2) y = f(kx + b)(k 6= 0) (3) y = f x 2 − 1 )µ (1) ✲ ✻ 0 π 4 π 2 3π 4 - π 4 - π 2 - 3π 4 x y (2) (k, b > 0)
9.若已知函数y=f(x)的图形,作函数v=|f(x),v=f(-x),3=-f(-x)的图形,并说明v,y,y的图形 与y的图形的关系 解:设y=f(x)的图形如下 则y的图形为 则y的图形为 则y的图形为 D的图形与(的图形类于与)的图断形关打2对称,当/(0)>0与的图形一样 y的图形与y的图形关于原点对称
13 ✲ ✻ 0 π−2b 2k π−b k 3π−2b 2k - b k - π+2b 2k - π+b k x y (3) ✲ ✻ 0 2 π π -π -2π x y 9. eƺÍy = f(x)„/ßäºÍy1 = |f(x)|, y2 = f(−x), y3 = −f(−x)„/ßø`²y1, y2, y3„/ Üy„/'X. )µy = f(x)„/Xeµ ✲ ✻ 0 x y Ky1„/èµ ✲ ✻ 0 x y Ky2„/èµ ✲ ✻ 0 x y Ky3„/èµ ✲ ✻ 0 x y y1„/f(x) < 0ûÜy„/'ux¶È°ßf(x) > 0ûÜy„/ò¶ y2„/Üy„/'uy¶È°ß y3„/Üy„/'u:È°ß
10.若已知f(x),9(x)的图形,试作函数y=5{f(x)+9(x)+|(x)-9(x)}的图形,并说明y的图形与∫(x),g(x)图 形的关系 解:y=max{f(x),9(x)} ∫(x) g(ar) 11.对于定义在[0,可上的函数y=x,先把它延拓到0,2可使它关于x=丌为对称,然后再把已延拓到0,2r]上的函 数延拓到整个实轴上使函数为以2为周期的函数 解:所求函数为:f(x)= 2r-x,x∈[r,2丌 x-2n丌,x∈[2n丌,(2n+1)r](n=±1,±2,…) 2nr-x,x∈[(2n-1)丌,2n](n=0,-1,±2,…) x+丌
14 10. eÆf(x), g(x)„/ߣäºÍy = 1 2 {f(x) + g(x) + |f(x) − g(x)|}„/ßø`²y„/Üf(x), g(x)„ /'X. )µ y = max{f(x), g(x)} ✲ ✻ 0 x y g(x) f(x) 11. Èu½¬3[0, π]˛ºÍy = xßkrßÚˇ[0, 2π]¶ß'ux = πèÈ°ß,2rÆÚˇ[0, 2π]˛º ÍÚˇá¢¶˛¶ºÍè±2π豜ºÍ. )µ§¶ºÍè:f(x) = x, x ∈ [0, π] 2π − x, x ∈ [π, 2π] x − 2nπ, x ∈ [2nπ,(2n + 1)π](n = ±1, ±2, · · ·) 2nπ − x, x ∈ [(2n − 1)π, 2nπ](n = 0, −1, ±2, · · ·) = π x π − 2 h x + π 2π i ✲ ✻ ❅ ❅ ❅ ❅❅ ❅ ❅ ❅ ❅❅ −π 0 π 2π -π -2π x y π
第二章极限与连续 §1.数列的极限和无穷大量 1.写出下列数列的前四项 (1)In =sinn (3)xn +…+ (4)n1=a>0,m=b>0,xn+1=Vn,+1=当 (5)x2n=1++…+(n=1,2,3 (1)x1=1sm1,x2=1sm.,x3=1im27,x4=1im64 (2)x1=m,x2=m(mn=12,x3=m(m=1m=2x2 x4=m(m=1m=2(m=3)x 均一++ (4)x1=a, Ig=vab, a yara+6 z4=Vab.(/a+6 y1= 次a+b (√a+ 3 (5)x2=1,x3=1,x4=,5=2 2.按定义证明以下数列为无穷小量 (3)2+(-1 (6)(-1)"(.99)0
15 1 Ÿ 4 Å Ü Î Y §1. Í 4 Å ⁄ à ° å ˛ 1. — e Í c o ë µ (1) x n = 13n sin n 3 (2) x n = m ( m − 1)· · · ( m − n + 1) n ! x n (3) x n = 1 √ n 2 + 1 + 1 √ n 2 + 2 + · · · + 1 √ n 2 + n (4) x 1 = a > 0, y 1 = b > 0, x n+1 = √ x n y n, y n+1 = x n + y n 2 (5) x 2 n = 1 + 12 + · · · + 1n ( n = 1 , 2 , 3 , · · · ) x 2 n+1 = 1n ( n = 1 , 2 , · · · ) ) : (1) x 1 = 13 sin 1, x 2 = 16 sin 8, x 3 = 19 sin 27, x 4 = 1 12 sin 64 (2) x 1 = mx, x 2 = m ( m − 1) 2 x 2 , x 3 = m ( m − 1)( m − 2) 6 x 3 , x 4 = m ( m − 1)( m − 2)( m − 3) 24 x 4 (3) x 1 = √1 2 , x 2 = 1√ 5 + 1√ 6 , x 3 = 1 √ 10 + 1 √ 11 + 1 √ 12 , x 4 = 1 √ 17 + 1 √ 18 + 1 √ 19 + 1 √ 20 (4) x 1 = a, x 2 = √ ab, x 3 = r √ ab a + b 2 , x 4 = √8 ab · 4r a + b 2 · √ a + √ b 2 y 1 = b, y 2 = a + b 2 , y 3 = ( √ a + √ b ) 2 4 , y 4 = ( √ a + √ b ) 2 4 + √4 ab p 2( a + b ) 16 (5) x 2 = 1, x 3 = 1, x 4 = 32 , x 5 = 12 2. U ½ ¬ y ² ± e Í èà ° ˛ µ (1) n + 1 n2 + 1 (2) sin n n (3) n + ( −1) n n 2 − 1 (4) 1n ! (5) 1n − 12n + 13n − · · · + ( −1) n+1 1n2 (6) ( −1) n (0 .999) n (7) 1n + e − n