13.(1)设f()=√1+x2(x>0),求f(x) (2)设∫(sin3)=1+c0sx,求f(cosx) 解(1)由f(-)=Ⅵh+x2,知f(x)=+ (2)f(sin)=1+cos x=2-2sin,: . f(x)=2-2x2, t f(cos x)=2-2cos" x=2sin'x 14.设∫(x)=8gnx={0.|=1及g(x)=e,求/[g(x)和gU(x)],并作出 两个函数的图形 解当x<0时g(x)=e2<1;,当x=0时,g(x=e=1,当x>0时,g(x 0, e2<1.故fg(x={0,x=0,同样易得g(x={1,1=1 e 图形见图1.3(a)和1.3(b) 图1.3(a) 图1.3(b) 15.证明 (1)shr+ shy= 2sh x+yx-y (2) shrchy=-Ish(x +y+sh(x-y)I x+y x+y x-y x-y x+ybx-) e- +e7 13. (1) 设 1 2 f xx ( ) 1 ( 0) x =+ > , 求 f ( ) x ; (2) 设 (sin ) 1 cos 2 x f = + x , 求 f (cos ) x . 解 (1) 由 1 2 f () 1 x x = + , 知 1 1 2 2 f () 1 ( ) 1 x x x x = + =+ . (2) 2 (sin ) 1 cos 2 2sin 2 2 x x f x =+ = − , ∴ 2 f () 2 2 x x = − , 故 f (cos ) x 2 = − 2 2cos x 2 = 2sin x . 14. 设 1, 1, ( ) sgn 0, 1, 1, 1 x fx x x x ⎧ < ⎪ == = ⎨ ⎪ − > ⎩ 及 () ex g x = , 求 f [ ( )] g x 和 g [ ( )] f x , 并作出 两个函数的图形. 解 当 x < 0 时, () e 1 x g x = < ; 当 x = 0 时, () e 1 x g x = = ; 当 x > 0 时, g( ) x e 1 x = < . 故 1, 0, [ ( )] 0, 0, 1, 0; x f gx x x ⎧ < ⎪ = = ⎨ ⎪ ⎩− > 同样易得 g[ ( )] f x e, 1, 1, 1, 1 , 1. e x x x ⎧ ⎪ < ⎪⎪ = ⎨ = ⎪ ⎪ > ⎪⎩ 图形见图 1.3(a)和 1.3(b). 15. 证明: (1) sh sh 2sh ch 2 2 x y xy x y + − + = ; (2) 1 sh ch [sh( ) sh( )] 2 x y xy xy = ++ − . 证 (1) 2 22 2 ee ee 2sh ch 2 22 2 2 x y xy xy xy xy xy + +− − − − +− − + =⋅ ⋅ O 1 −1 x 图 1.3(a) y −1 O 1 x 1 e 1 e 图 1.3(b) y