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+29 §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