(1)设0≤x1<x2 则y2-y1=2-x1=(x2+x1)(x2-x1)>0,于是函数y=x2当0≤x时严格单调增加 (2)设-7≤x1<x2≤ SWUy2-n=sin z2-sin z1 =2 cos 22+Ii sin 12-1 又2≤n1<2≤7则+m2∠,0<2 是 x1+x2 0 0,从而y->0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(-∞<x≤0) (2)y=cosx(0≤x≤丌) 证明 (1)设0≤x1<x2 则y2-y=x2-x2=(x2+x1)(x2-x1)<0,于是函数y=x2当x≤O时严格单调减少 (2)设0≤x1<x2≤丌 又0≤1<2≤丌,则<+22如空一要 UJy2-yI cos x2-cos11=-2sin 2T 25x0<2≤x,于是n2+2>0.m>0,从 而y-<0即函数y=cosx当0≤x≤π时严格单调减少 21.讨论下列函数的奇偶性 (1)y=x+x2-x5 (2)y=a+boost (3)y=r+sinr +e (4)y=rsin 当x>0时 5)y=sgmx=0,当x=0时 1 <<+ 当-∞<x 时 (1)因y=f(x)=x+x2-x5,则f(-x)=-x+x2+x5,故f(-x)≠f(x),f(-x)≠-f(x),于是此函数 是非奇非偶函数 (2)因y=f(x)=a+ bcos ar,则f(-x)=a+bcos(-x)=a+ bcos x=∫(x),于是此函数是偶函数 (3)因y=f(x)=x+sinx+e2,则f(-x)=-x-sinx+e-,故f(-x)≠∫(x),f(-a)≠-f(x),于是此 函数是非奇非偶函数 (4)因y=f(x)=xsin,则f(-)=-r出n=x=xnx=f(x),于是此函数是偶函数 1,当x>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 < x2 Ky2 − y1 = x 2 2 − x 2 1 = (x2 + x1)(x2 − x1) > 0ßu¥ºÍy = x 20 6 xûÓǸNO\. (2) − π 2 6 x1 < x2 6 π 2 Ky2 − y1 = sin x2 − sin x1 = 2 cos x2 + x1 2 sin x2 − x1 2 q− π 2 6 x1 < x2 6 π 2 ßK− π 2 < x1 + x2 2 < π 2 , 0 < x2 x1 2 6 π 2 ßu¥cos x1 + x2 2 > 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 (−∞ < x 6 0) (2) y = cos x(0 6 x 6 π) y²µ (1) 0 6 x1 < x2 Ky2 − y1 = x 2 2 − x 2 1 = (x2 + x1)(x2 − x1) < 0ßu¥ºÍy = x 2x 6 0ûÓǸN~. (2) 0 6 x1 < x2 6 π Ky2 − y1 = cos x2 − cos x1 = −2 sin x2 + x1 2 sin x2 − x1 2 q0 6 x1 < x2 6 πßK0 < x1 + x2 2 < π, 0 < x2 x1 2 6 π 2 ßu¥sin x1 + x2 2 > 0, sin x2 − x1 2 > 0ßl y2 − y1 < 0=ºÍy = cos x0 6 x 6 πûÓǸN~. 21. ?ÿeºÍ¤Û5µ (1) y = x + x 2 − x 5 (2) y = a + b cos x (3) y = x + sin x + e x (4) y = x sin 1 x (5) y = sgnx = 1, x > 0û 0, x = 0û −1 x < 0û (6) y = 2 x2 , 1 2 < x < +∞û sin x 2 , − 1 2 6 x 6 1 2 û 1 2 x 2 , − ∞ < x < − 1 2 û )µ (1) œy = f(x) = x + x 2 − x 5ßKf(−x) = −x + x 2 + x 5ßf(−x) 6= f(x), f(−x) 6= −f(x)ßu¥dºÍ ¥ö¤öÛºÍ. (2) œy = f(x) = a + b cos xßKf(−x) = a + b cos(−x) = a + b cos x = f(x)ßu¥dºÍ¥ÛºÍ. (3) œy = f(x) = x + sin x + e xßKf(−x) = −x − sin x + e −xßf(−x) 6= f(x), f(−x) 6= −f(x)ßu¥d ºÍ¥ö¤öÛºÍ. (4) œy = f(x) = x sin 1 x ßKf(−x) = −x sin 1 −x = x sin 1 x = f(x)ßu¥dºÍ¥ÛºÍ. (5) œy = f(x) = 1, x > 0û 0, x = 0û −1 x < 0û ß Kf(−x) = 1, − x > 0û 0, − x = 0û −1 − 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 < x < +∞û sin x 2 , − 1 2 6 x 6 1 2 û 1 2 x 2 , − ∞ < x < − 1 2 û ß Kf(−x) = 2 (−x) 2 , 1 2 < −x < +∞û sin(−x) 2 , − 1 2 6 −x 6 1 2 û 1 2 (−x) 2 , − ∞ < −x < − 1 2 û = 1 2 x 2 , 1 2 < x < +∞û sin x 2 , − 1 2 6 x 6 1 2 û 2 x2 , − ∞ < x < − 1 2 û ß f(−x) 6= f(x), f(−x) 6= −f(x)ßu¥dºÍ¥ö¤öÛºÍ. 22. £y¸áۺͶ»¥ÛºÍ߸᤺Ͷ»¥¤ºÍßò᤺ÍÜòáۺͶ»¥¤ºÍ. y²µf1(x), f2(x)转3(−a, a)(a > 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