复旦大学：《数学分析》经典教材的课后习题答案

第一篇极限论 第一部分极限初论 第一章变量与函数 §1.函数的概念 1.解下列不等式,并画出x的范围: (1)-2-3 (2)10时,x1+ 当a<0时,1+<x<1 当a=0时,x<1

1 1òü 4Åÿ 1ò‹© 4Å–ÿ 1òŸ C˛ÜºÍ §1. ºÍVg 1. )eÿ™ßøx—xâåµ (1) −2 0 )µ (1) x − 3 2 ✘✛ ✲ -1-2-3 0 x ❜❜ (2) 1 0ûßx 1 + 1 a ¶ ✘✛ ✲ 0 1 x 1 + 1 a ❝ ❝ a < 0ûß1 + 1 a < x < 1 ✛ ✘ ✲ 0 1 x 1 + 1 a ❝❝ a = 0ûßx < 1 ✘ ✲ 0 1 x ❝

(4)2kx+≤x≤2k丌+或2kx-≤x≤2kx-(k∈Z) (5)-4|x+1 (2)2A (4)|x-a0 x+2 (1)x< (2)-2<x

2 (4) 2kπ + π 3 6 x 6 2kπ + π 2 ½2kπ − π 2 6 x 6 2kπ − π 3 (k ∈ Z) ✲ ✄ ￾ ✄ ￾ ✄ ￾ ✄ ￾ 0 x (5) −4 ||x| − |y|| (2) |x1 + x2 + x3 + · · · + xn| 6 |x1| + |x2| + · · · + |xn| (3) |x + x1 + · · · + xn| > |x| − (|x1| + · · · + |xn|) y²µ (1) œ|x||y| > xyßK(x − y) 2 > (|x| − |y|) 2ßu¥|x − y| > ||x| − |y|| (2) ^ÍÆ8B{y². (i) n = 2ûßd|x1 + x2| 6 |x1| + |x2|ß(ÿ§·. (ii) bn = kû(ÿ§·ß=k|x1 + x2 + x3 + · · · + xk| 6 |x1| + |x2| + · · · + |xk|. Kn = k + 1ûß|x1 + x2 + x3 + · · · + xk+1| 6 |x1 + x2 + x3 + · · · + xk| + |xk+1| 6 |x1| + |x2| + · · · + |xk| + |xk+1| n˛åßÈòÉg,Ínß|x1 + x2 + x3 + · · · + xn| 6 |x1| + |x2| + · · · + |xn|˛§·. (3) |x + x1 + · · · + xn| > |x| − |x1 + x2 + x3 + · · · + xn| > |x| − (|x1| + · · · + |xn|) 3. )e˝Èäÿ™ßøx—xâåµ (1) |x| > |x + 1| (2) 2 A (4) |x − a| 0 (5)     x − 2 x + 1     > x − 2 x + 1 (6) 2 < 1 |x + 2| < 3 )µ (1) x < − 1 2 ✘ ✲ -1 0 x ❜ (2) − 1 2 < x < − 1 4 ½ 1 4 < x < 1 2 ✲ ☛ ✟ ☛ ✟ 0 1 x 2 - 1 2 ❡❡ ❡ ❡

(3)当A≥0时,xA 当A<0时,x∈R (4)a-n<a<a+n (5)原式等价于x-2∠0,则-1<x<2 <x< <x<一一 4.求下列函数的定义域及它在给定点上的函数值 (1)y=f(x)=-x+-的定义域及f(-1),f(1)和f(2) (2)y=/()=√a2的定义域及/(,J(和(-2) 1 (3)=8()=c的定义域及(1),(2 (4)y=g(a)=a2tana的定义域及g(0),g ()x=2()2=sm0+c的定义域及x(-2),x(-m) (6)y=f(x)= 的定义域及f(0),f(-1)

3 (3) A > 0ûßx A ✘ ✛✲ 0 A-A x ❡❡ A < 0ûßx ∈ R (4) a − η < x < a + η ✛a ✘ ✲ − η 0 a + ηa x ❡❡ (5) ™du x − 2 x + 1 < 0ßK−1 < x < 2 ✛ ✘✲ -1 1 2 0 x ❜ ❜ (6) − 5 3 < x < − 3 2 ½− 5 2 < x < − 7 3 ✲ ✞ ☎ ✞ ☎ -1-2-3 0 x ❡❡ ❡❡ 4. ¶eºÍ½¬ç9ß3â½:˛ºÍäµ (1) y = f(x) = −x + 1 x ½¬ç9f(−1), f(1)⁄f(2)¶ (2) y = f(x) = √ a 2 − x2½¬ç9f(0), f(a)⁄f  − a 2  ¶ (3) s = s(t) = 1 t e −t½¬ç9s(1), s(2)¶ (4) y = g(α) = α 2 tan Ὤç9g(0), g  π 4  , g  − π 4  ¶ (5) x = x(θ) = sin θ + cos 转ç9x  − π 2  , x(−π) (6) y = f(x) = 1 (x − 1)(x + 2)½¬ç9f(0), f(−1)

(1)函数的定义域为X=(-∞,O)U(0,∞),f(-1)=0.f(1)=0,f(2)= (2)函数的定义域为x=-al,0)=la)=0(-2) (3)函数的定义域为(-∞,0)U0,∞x),(1)=1,(2)=2 (函数的定义域为{∈Bx≠k+三,k∈2,00=09(理) 426 )函数的定义域为X=(一×,∞)x(-2)=-1x(-m)=-1 ()函数的定义域为X=(-x,-2)U(-2.1)U.+∞),f0=-2,(-1)=-2 5.求下列函数的定义域及值域: (1)y=√2+x-x2 (2)y=Cost (3)y=In(sin (4)y=ainπr (1)函数的定义域为X=[-1,2,值域为/O3 (2)函数的定义域为2k丌-,2k丌+5(k∈Z),值域为,1 (3)函数的定义域为(11 2k+12k)(k∈Z,值域为(-∞ (4)函数的定义域为(n-1,n)(n=0,±1,±2,……),值域为(-∞,-1]U[1,+∞) 6.设f(x)=x+1,y(x)=x-2,试解方程f(x)+(x)=|f(x)+|p(x) 解:由已知,得∫(x)y(x)≥0即(x+1)(x-2)≥0,则x≥2或x≤-1 7.设f(x)=(|x+x)(1-x),求满足下列各式的x值: (1)f(0)=0 (2)f(x)1 8.图1-5表示电池组V、固定电阻R和可变电阻R组成的电路在一段不长的时间内,A,B两点间的电压V可以看 成一个常量求出电流Ⅰ和可变电阻R的函数式 解:由已知及物理学知识,得V=I(R0+R) 9.在一个圆柱形容器内倒进某种溶液,该圆柱形容器的底半径是a,高为h,倒进溶液的高度是x(图1-6).该 溶液的容积V和x之间的函数关系V=V(x),并写出它的定义域和值域 解:由已知,得V=a2x,它的定义域为[0,,值域为[1,ma2h 10.某灌溉渠的截面积是一个梯形,如图1-7,底宽2米,斜边的倾角为45°,CD表示水面,求截面ABCD的面 积S与水深h的函数关系 解:由已知及图,得S=h(h+2) 11.有一深为H的矿井,如用半径为R的卷扬机以每秒钟弧度的角速度从矿井内起吊重物,求重物底面与地面的 距离s和时间t的函数关系(图1-8) 已知及图,得s=H 1+x2,x0.求f(-2,f(-1)(0()和(2 由已如1得23=50)=210=-10-0()-

4 )µ (1) ºÍ½¬çèX = (−∞, 0) S (0,∞)ßf(−1) = 0, f(1) = 0, f(2) = − 3 2 (2) ºÍ½¬çèX = [−|a|, |a|]ßf(0) = |a|, f(a) = 0, f  − a 2  = √ 3 2 |a| (3) ºÍ½¬çè(−∞, 0) S (0,∞)ßs(1) = 1 e , s(2) = 1 2e 2 (4) ºÍ½¬çè n x   x ∈ R, x 6= kπ + π 2 , k ∈ Z o ßg(0) = 0, g  π 4  = π 2 16 , g  − π 4  = − π 2 16 (5) ºÍ½¬çèX = (−∞,∞)ßx  − π 2  = −1, x(−π) = −1 (6) ºÍ½¬çèX = (−∞, −2) S (−2, 1) S (1, +∞)ßf(0) = − 1 2 , f(−1) = − 1 2 5. ¶eºÍ½¬ç9äçµ (1) y = √ 2 + x − x2 (2) y = √ cos x (3) y = ln  sin π x  (4) y = 1 sin πx )µ (1) ºÍ½¬çèX = [−1, 2]ßäçè  0, 3 2  (2) ºÍ½¬çè h 2kπ − π 2 , 2kπ + π 2 i (k ∈ Z)ßäçè[0, 1] (3) ºÍ½¬çè  1 2k + 1 , 1 2k  (k ∈ Z)ßäçè(−∞, 0] (4) ºÍ½¬çè(n − 1, n)(n = 0, ±1, ±2, · · ·)ßäçè(−∞, −1] S [1, +∞) 6. f(x) = x + 1, ϕ(x) = x − 2ߣ)êß|f(x) + ϕ(x)| = |f(x) + |ϕ(x)| )µdÆßf(x)ϕ(x) > 0=(x + 1)(x − 2) > 0ßKx > 2½x 6 −1. 7. f(x) = (|x| + x)(1 − x)߶˜veà™xäµ (1) f(0) = 0 (2) f(x) 0ßKáf(x) 1 8. „1-5L´>³|V !½>{R0⁄åC>{R|§>¥.3ò„ÿûmSßA, B¸:m>ÿV å±w §òá~˛.¶—>6I⁄åC>{RºÍ™. )µdÆ9‘nÆ£ßV = I(R0 + R). 9. 3òáŒ/NÏS?,´MóßTŒ/NÏ.媥aßpèhß?Móp›¥x£„1-6§. T MóN»V ⁄xÉmºÍ'XV = V (x)ßø—ß½¬ç⁄äç. )µdÆßV = πa2xßß½¬çè[0, h]ßäçè[1, πa2h] 10. ,/Y±°»¥òáF/ßX„1-7ß.°2íß>ñè45oßCDL´Y°ß¶°ABCD° »SÜYhºÍ'X. )µdÆ9„ßS = h(h + 2). 11. kòèH¶³ßX^åªèRÚűz¶®ωl›Ñ›l¶³SÂL­‘߶­‘.°Ü/° Âls⁄ûmtºÍ'X£„1-8§. )µdÆ9„ßs = H − ωRt  t ∈  0, H ωt 12. y = f(x) =  1 + x 2 , x 0 ߶f(−2), f(−1), f(0), f(1)⁄f  1 2  . )µdÆßf(−2) = 5, f(−1) = 2, f(0) = −1, f(1) = 0, f  1 2  = − 1 2

0≤tD即(2)+fe2)> (2)对于所有的x,2(1≠n)成立 ∫(x) 9.证明下列各函数在所示区间内是单调增加的函数 (1)y=x2(0≤x<+∞) (2)y=sinr(≤x≤需 证明

5 13. x(t) =    0, 0 6 t f  x1 + x2 2  È u§kx1, x2(x1 6= x2)§·£Ö‰k˛„A5ºÍ⇺ͧ. y²µ3­Ç˛?¸:A(x1, f(x1)), B(x2, f(x2))ßÎABߟ•:C(xC , yC )ßKf(x1) + f(x2) = 2yC , x1 + x2 = 2xC q­Ç˛xD = x1 + x2 2 §È:pãIèyD = f  x1 + x2 2  ßKxC = xD q­Çy = f(x)˛?ò^u—puߧÅlÖx1, x2èuÜl:ßKyC > yD= f(x1) + f(x2) 2 > f  x1 + x2 2  Èu§kx1, x2(x1 6= x2)§·. ✲ ✻ ✜ ✜ ✜ ✜ ✜ ✜ ✜ 0 x1 x2 x A C B xD y f(x) 19. y²eàºÍ3§´´mS¥¸NO\ºÍµ (1) y = x 2 (0 6 x < +∞) (2) y = sin x  − π 2 6 x 6 π 2  y²µ

(1)设0≤x10,于是函数y=x2当0≤x时严格单调增加 (2)设-7≤x10即函数y=sinx当一-≤x≤x时严格单调增加 20.证明下列函数在所示区间内是单调减少的函数 (1)y=x2(-∞0.m>0,从 而y-0时 5)y=sgmx=0,当x=0时 1 0时 (5)因y=f(x)=0.当x=0时 0时 1,当x>0时 则f(-x)={0,当-x=0时=0.当x=0时=-f(),于是此函数是奇函数 1当一x<0时 1当x<0时

6 (1) 0 6 x1 0ßu¥ºÍy = x 20 6 xûÓǸNO\. (2) − π 2 6 x1 0, sin x2 − x1 2 > 0ßl y2 − y1 > 0=ºÍy = sin x− π 2 6 x 6 π 2 ûÓǸNO\. 20. y²eºÍ3§´´mS¥¸N~ºÍµ (1) y = x 2 (−∞ 0, sin x2 − x1 2 > 0ßl y2 − y1 0û 0, x = 0û −1 x 0û 0, x = 0û −1 x 0û 0,  − x = 0û −1  − x 0û 0, x = 0û 1 x < 0û = −f(x)ßu¥dºÍ¥¤ºÍ

7 x2,当-≤x≤时 则f(-x) (-x)2,当-5≤-x≤时 故f(-x)≠f(x),f(-x)≠-f(x),于是此函数是非奇非偶函数 2.试证两个偶函数的乘积是偶函数,两个奇函数的乘积是奇函数,一个奇函数与一个偶函数的乘积是奇函数 证明:设f1(x),f2(x)为定义在(-a,a)(a>0)内的偶函数,g1(x),92(x)为定义在(-a,a)(a>0)内的奇函 数,F1(x)=f1(x)f2(x),F2(x)=g1(x)92(x),F3(x)=f1(x)f2(x) 则f1(-x)=f1(x),f2(-x)=f2(x),g1(x)=-g1(x),92(-x)=-92(x),于是 F1(-x)=f1(-x)f2(-x)=f1(x)(x)=F1(x) F2(-x)=91(-)92(-x)=(-91(x)(-92(x)=91(x)92(x)=F2(x) F3(-x)=f1(-x)91(-x)=f1(x)(-91(x)=-f1(x)g1(x)=-F3(x) 从而F1(x)是偶函数;F2(x)是偶函数;F3(x)是奇函数 23.设f(x)为定义在(-∞,+∞)内的任何函数,证明F1(x)≡=f(x)+f(-x)是偶函数,F2(x)≡f(x)-f(-x)是奇 函数写出对应于下列函数的F1(x),F2(x) (2)y=(1+x) 证明:因F1(-x)=f(-x)+f(x)=F1(x),则F1(x)=∫(x)+f(-x)是偶函数 又F2(-x)=f(-x)-f(x)=-F2(x),则F2(x)=f(x)-f(-x)是奇函数 (1)F1(x)=f(x)+f(-x)=a2+a-x,F2(x)=f(x)-f(-x)=a2 (2)F1(x)=f(x)+f(-x)=(1+x)2+(1-x)n,F2(x)=f(x)-f(-x)=(1+x)-(1-x)n 4.说明下列函数哪些是周期函数,并求最小周期: (1)y=sina (2)y=sinr (3)y=sin r +o sin 2r (5)y=lsin r|+Icos rl (6)y=√tanx (7)y=x-[x (8)y=sinner (1)因=出n2s1os2x,则r= 22 (2)假设y=sinx2为一周期函数且T=>0 据周期函数的定义,对任何x∈(-∞,+∞),有sin(x+u)2=sinx2,特别对x=0也应该成立 则sinu2=0,于是u2=kr,u=Vkr(k∈z+) 又对x=√2u=√2也成立,故sin(√②u+u)2=sin2=0,则(√2+1)2kr=n丌(n∈z+),于 是(√2+1)2=(k,n∈z+) 又(√2+1)2=3+2V2∈Q,而∈Q+,则假设不成立,即函数y=sinx2不是周期函数 (3)因n=smz的T=27:=2sin2的T=,则y=inx+2sm2的T=27 (4)T=元=8

7 (6) œy = f(x) =    2 x2 ,  1 2 0)SÛºÍßg1(x), g2(x)转3(−a, a)(a > 0)S¤º ÍßF1(x) = f1(x)f2(x), F2(x) = g1(x)g2(x), F3(x) = f1(x)f2(x) Kf1(−x) = f1(x), f2(−x) = f2(x), g1(x) = −g1(x), g2(−x) = −g2(x)ßu¥ F1(−x) = f1(−x)f2(−x) = f1(x)f2(x) = F1(x) F2(−x) = g1(−x)g2(−x) = (−g1(x))(−g2(x)) = g1(x)g2(x) = F2(x) F3(−x) = f1(−x)g1(−x) = f1(x)(−g1(x)) = −f1(x)g1(x) = −F3(x) l F1(x)¥ÛºÍ¶F2(x)¥ÛºÍ¶F3(x)¥¤ºÍ. 23. f(x)转3(−∞, +∞)S?¤ºÍßy²F1(x) ≡ f(x) + f(−x)¥ÛºÍßF2(x) ≡ f(x) − f(−x)¥¤ ºÍ.—ÈAueºÍF1(x), F2(x)µ (1) y = a x (2) y = (1 + x) n y²µœF1(−x) = f(−x) + f(x) = F1(x)ßKF1(x) = f(x) + f(−x)¥ÛºÍ qF2(−x) = f(−x) − f(x) = −F2(x)ßKF2(x) = f(x) − f(−x)¥¤ºÍ. (1) F1(x) = f(x) + f(−x) = a x + a −x , F2(x) = f(x) − f(−x) = a x − a −x (2) F1(x) = f(x) + f(−x) = (1 + x) n + (1 − x) n , F2(x) = f(x) − f(−x) = (1 + x) n − (1 − x) n 24. `²eºÍ= ¥±œºÍßø¶Å±œµ (1) y = sin2 x (2) y = sin x 2 (3) y = sin x + 1 2 sin 2x (4) y = cos π 4 x (5) y = | sin x| + | cos x| (6) y = √ tan x (7) y = x − [x] (8) y = sin nπx )µ (1) œy = sin2 x = 1 2 − 1 2 cos 2xßKT = 2π 2 = π (2) by = sin x 2èò±œºÍÖT = ω > 0 ‚±œºÍ½¬ßÈ?¤x ∈ (−∞, +∞)ßksin(x + ω) 2 = sin x 2ßAOÈx = 0èAT§·ß Ksin ω 2 = 0ßu¥ω 2 = kπ, ω = √ kπ(k ∈ Z +) qÈx = √ 2ω = √ 2kπ觷ßsin(√ 2ω + ω) 2 = sin ω 2 = 0ßK( √ 2 + 1)2 kπ = nπ(n ∈ Z +)ßu ¥( √ 2 + 1)2 = k n (k, n ∈ Z + ) q( √ 2 + 1)2 = 3 + 2√ 2 ∈ Q −ß k n ∈ Q +ßKbÿ§·ß=ºÍy = sin x 2ÿ¥±œºÍ. (3) œy1 = sin xT = 2π¶y2 = 1 2 sin 2xT = πßKy = sin x + 1 2 sin 2xT = 2π. (4) T = 2π π 4 = 8

()因(2)=1m+1/(x+2)=m(+2)+∞(x+32)=1+1sm=/) 据经验,知y =| sin el+|cosx的T= (6)因f(x)=tanx的T=丌,则y= Tana的T=丌 (7)因y=x-团]=(x),则y=x-[]的T=1

8 (5) œf(x) = | sin x| + | cos x|, f  x + π 2  =    sin  x + π 2    +   cos  x + π 2    = | cos x| + | sin x| = f(x) ‚²ßy = | sin x| + | cos x|T = π 2 . (6) œf(x) = tan xT = πßKy = √ tan xT = π. (7) œy = x − [x] = (x)ßKy = x − [x]T = 1. (8) T = 2π nπ = 2 n

9 复合函数和反函数 1.下列函数能否构成复合函数y=∫(φ(x),如果能够构成则指出此复合函数的定义域和值域: (1)y=f(u)=2,u=p(x)=x2 (2)y=f(u)=lnu,u=y(x)=1-x2 (3)y=f(u)=u2+u3,u=g(x) 1,当x为有理数时 1,当x为无理数时 (4)y=∫(u)=2,定义域为U,u=φ(x),定义域为X,值域为U2 (5)y=f(u)=Vu,u=p(a)=cos r (1)因y=f(a)=2的定义域为(-∞,+∞),u=y(x)=x2的值域为[0,+∞) 则此函数能构成复合函数y=22,它的定义域为(-∞,+∞),值域为[1,+x) 2)因y=f(u)=lnu的定义域为(0,+∞),u=g(x)=1-x2的值域为(-∞,1] 则此函数能构成复合函数y=ln(1-x2),它的定义域为(-1,1),值域为(-∞,0 (3)因y=f(u)=2+u3的定义域为(-∞,+∞), 1,当x为有理数 u=(x)={-1,当z为无理数时的值域为{-1,1} 则此函数能构成复合函数y=「2,当x为有理数时,它的定义域为(-∞,+∞),值域为{02} 当x为无理数时 (4)因y=f(u)=2的定义域为U1,u=(x)的值域为U2 当Un∩U2≠φ时,此函数能构成复合函数y=2,它的定义域视具体函数而定,值域为{2} 当U∩U2=φ时,此函数不能构成复合函数 (5)因y=∫(u)=√u的定义域为0,+∞),=y(x)=cosr的值域为[-1,1 则此函数能构成复合函数y=√@x,它的定义域为[2kx-5,2kx+(k=0±1,+2…),值域 为0,1 2.设∫(x)=ax2+bx+c,证明f(x+3)-3f(x+2)+3f(x+1)-f(x)≡0 证明:由已知,得 f(x+3)-3f(x+2)+3f(x+1)-f(x)=a(x+3)2+b(x+3)+c-3a(x+2)2+b(x+2)+d+3{a(x+1)2+ b(x+1)+-(ax2+b+e)=a[(x+3)2-x2]+b(x+3-x)-3a[(x+2)2-(x+1)2]-3bx+2-(x+1) ar+9a+3b-3a(2x+3)-3b≡0 (2)设y=f(x)=x2ln(1+x),求f(e) (3)设y=∫(x)=√1+x+x2,求f(x2)及f(-x2) (4)设y=m(=~1 求f( a tan r) x+2 (2)因y=f(x)=x2ln(1+x),则f(e-)=(e-)2ln(1+e)= 则f(x2)=√1+x2+x,f(-x2)=√1-x2+x (4)因y=f(t) 则f( a tan o) 4.若∫(x)=x2,y(x)=2,求f((x)及p(f(x) 解:因(x)=x2,y(x)=2,则f(y(x)=(2)2=2=4,p(f(x)=2 解:因y(x)=x3+1,则 (x2)=(x2)3+1=x°6+1,(y(x)2=(x3+1)2=x6+2x3+1,y(yp(x)=(x32+1)3+1=x+3x°+3x3+2

9 §2. E‹ºÍ⁄áºÍ 1. eºÍUƒ§E‹ºÍy = f(ϕ(x))ßXJU §Kç—dE‹ºÍ½¬ç⁄äçµ (1) y = f(u) = 2u , u = ϕ(x) = x 2 (2) y = f(u) = ln u, u = ϕ(x) = 1 − x 2 (3) y = f(u) = u 2 + u 3 , u = ϕ(x) =  1, xèknÍû −1, xèÃnÍû (4) y = f(u) = 2ß½¬çèU1ßu = ϕ(x)ß½¬çèXßäçèU2 (5) y = f(u) = √ u, u = ϕ(x) = cos x )µ (1) œy = f(u) = 2u½¬çè(−∞, +∞)ßu = ϕ(x) = x 2äçè[0, +∞) KdºÍU§E‹ºÍy = 2x 2ßß½¬çè(−∞, +∞)ßäçè[1, +∞) (2) œy = f(u) = ln u½¬çè(0, +∞)ßu = ϕ(x) = 1 − x 2äçè(−∞, 1] KdºÍU§E‹ºÍy = ln(1 − x 2 )ßß½¬çè(−1, 1)ßäçè(−∞, 0] (3) œy = f(u) = u 2 + u 3½¬çè(−∞, +∞)ß u = ϕ(x) =  1, xèknÍû −1, xèÃnÍû äçè{−1, 1} KdºÍU§E‹ºÍy =  2, xèknÍû 0, xèÃnÍû ßß½¬çè(−∞, +∞)ßäçè{0, 2} (4) œy = f(u) = 2½¬çèU1ßu = ϕ(x)äçèU2 U1 T U2 6= φûßdºÍU§E‹ºÍy = 2ßß½¬ç¿‰NºÍ ½ßäçè{2}¶ U1 T U2 = φûßdºÍÿU§E‹ºÍ (5) œy = f(u) = √ u½¬çè[0, +∞)ßu = ϕ(x) = cos xäçè[−1, 1] KdºÍU§E‹ºÍy = √ cos xßß½¬çè h 2kπ − π 2 , 2kπ + π 2 i (k = 0, ±1, ±2, · · ·)ßäç è[0, 1] 2. f(x) = ax2 + bx + cßy²f(x + 3) − 3f(x + 2) + 3f(x + 1) − f(x) ≡ 0 y²µdÆß f(x+ 3)−3f(x+ 2) + 3f(x+ 1)−f(x) = a(x+ 3)2 +b(x+ 3) +c−3[a(x+ 2)2 +b(x+ 2) +c] + 3[a(x+ 1)2 + b(x+ 1) +c]−(ax2 +bx+c) = a[(x+ 3)2 −x 2 ] +b(x+ 3−x)−3a[(x+ 2)2 −(x+ 1)2 ]−3b[x+ 2−(x+ 1)] = 6ax + 9a + 3b − 3a(2x + 3) − 3b ≡ 0 3. (1) y = f(x) = a + bx + c x ߶f  2 x  (2) y = f(x) = x 2 ln(1 + x)߶f(e −x ) (3) y = f(x) = √ 1 + x + x2߶f(x 2 )9f(−x 2 ) (4) y = f(t) = 1 √ a 2 + x2 ߶f(a tan x) )µ (1) œy = f(x) = a + bx + c x ßKf  2 x  = a + 2b x + c 2 x = a + 2b x + cx 2 = cx2 + 2ax + 4b 2x (2) œy = f(x) = x 2 ln(1 + x)ßKf(e −x ) = (e −x ) 2 ln(1 + e −x ) = ln(e x + 1) − x e 2x (3) œy = f(x) = √ 1 + x + x2ßKf(x 2 ) = √ 1 + x2 + x4, f(−x 2 ) = √ 1 − x2 + x4 (4) œy = f(t) = 1 √ a 2 + x2 ßKf(a tan x) = 1 p a 2 + (a tan x) 2 = 1 √ a 2 sec2 x = 1 |a sec x| 4. ef(x) = x 2 , ϕ(x) = 2x߶f(ϕ(x))9ϕ(f(x)). )µœf(x) = x 2 , ϕ(x) = 2xßKf(ϕ(x)) = (2x ) 2 = 22x = 4x , ϕ(f(x)) = 2x 2 5. eϕ(x) = x 3 + 1߶ϕ(x 2 ),(ϕ(x))29ϕ(ϕ(x)). )µœϕ(x) = x 3 + 1ßK ϕ(x 2 ) = (x 2 ) 3 + 1 = x 6 + 1,(ϕ(x))2 = (x 3 + 1)2 = x 6 + 2x 3 + 1, ϕ(ϕ(x)) = (x 3 + 1)3 + 1 = x 9 + 3x 6 + 3x 3 + 2

6.设∫(x) 求f(f(x),f(f(f(x),f 因f(x)=,,则 f((r)= f(f((x))= 2=x,1 1 7.求下列函数的反函数及反函数的定义域 (1)y=x2(-∞<x≤0) (2)y=√1-x2(-1≤x≤0) 当-∞<x<1时 (4)y= 22,当4<x<+∞时 (1)因y=x2(-0<x≤0),则x=-√0≤y<+∞),从而此函数的反函数为y=-√(0≤y<+∞) (2)因y=√1-x2(-1≤x≤0),则x=-√1-y2(0≤y≤1),从而此函数的反函数为y=-√1-x2(0≤ x≤1) (3)因y=sinx(a≤x≤ 则x=丌- arcsin y(-1≤y≤1),从而此函数的反函数为y=r arcsin:L(-1≤x≤1) 当-∞<x<1时 <y<1 (4)因y={x,当1≤x≤4时 则x={√当1≤y≤16时,从而此函数的反函数 22,当4<x<+∞时 og2y,当16<x<+∞时 当-∞<x<1时 为y=1og2x,316<x<+∞时 当1≤x≤16时

10 6. f(x) = 1 1 − x ߶f(f(x)), f(f(f(x))), f  1 f(x)  . )µœf(x) = 1 1 − x ßK f(f(x)) = 1 1 − 1 1 − x = x − 1 x , f(f(f(x))) = 1 1 − 1 1 − 1 1 − x = 1 1 − x − 1 x = x, f  1 f(x)  = 1 1 − (1 − x) = 1 x 7. ¶eºÍáºÍ9áºÍ½¬çµ (1) y = x 2 (−∞ < x 6 0) (2) y = √ 1 − x2(−1 6 x 6 0) (3) y = sin x  π 2 6 x 6 3 2 π  (4) y =    x,  − ∞ < x < 1û x 2 , 1 6 x 6 4û 2 x , 4 < x < +∞û )µ (1) œy = x 2 (−∞ < x 6 0)ßKx = − √y(0 6 y < +∞)ßl dºÍáºÍèy = − √ x(0 6 y < +∞) (2) œy = √ 1 − x2(−1 6 x 6 0)ßKx = − p 1 − y 2(0 6 y 6 1)ßl dºÍáºÍèy = − √ 1 − x2(0 6 x 6 1) (3) œy = sin x  π 2 6 x 6 3 2 π  ßKx = π − arcsin y(−1 6 y 6 1)ßl dºÍáºÍèy = π − arcsin x(−1 6 x 6 1) (4) œy =    x,  − ∞ < x < 1û x 2 , 1 6 x 6 4û 2 x , 4 < x < +∞û ßKx =    y,  − ∞ < y < 1û √y, 1 6 y 6 16û log2 y, 16 < x < +∞û ßl dºÍáºÍ èy =    x,  − ∞ < x < 1û √ x, 1 6 x 6 16û log2 x, 16 < x < +∞û