解答:解:(Ⅰ)當(dāng)m=1時(shí),
f(x)=.
因?yàn)?span id="p5dvms6" class="MathJye">f′(x)=
,所以
k=f′()=.
因?yàn)?span id="7er52x7" class="MathJye">f(
)=
,所以函數(shù)f(x)在點(diǎn)
(,f())處的切線方程為12x-25y+4=0.…(6分)
(Ⅱ)
f′(x)=(2-m)(x2+m)-(2-m)x•2x |
(x2+m)2 |
=
(1)當(dāng)m=0時(shí),
f(x)=.
因?yàn)?span id="uubsp5s" class="MathJye">f′(x)=-
,當(dāng)f'(x)<0時(shí),x<0,或x>0.
所以函數(shù)f(x)的單調(diào)減區(qū)間為(-∞,0),(0,+∞),無單調(diào)增區(qū)間.
(2)當(dāng)m<0時(shí),f(x)的定義域?yàn)?span id="z0htl08" class="MathJye">{x|x≠±
}.
當(dāng)f'(x)<0時(shí),
x<-或-<x<或x>,
所以函數(shù)f(x)的單調(diào)減區(qū)間為
(-∞,-),(-,),(,+∞),無單調(diào)增區(qū)間.
(3)當(dāng)m>0時(shí),
f′(x)=.
①當(dāng)0<m<2時(shí),
若f'(x)<0,則
x<-或x>,
若f'(x)>0,則
-<x<,
所以函數(shù)f(x)的單調(diào)減區(qū)間為
(-∞,-),(,+∞),
函數(shù)f(x)的單調(diào)增區(qū)間為
(-,).
②當(dāng)m=2時(shí),f(x)=0,為常數(shù)函數(shù),無單調(diào)區(qū)間.
③當(dāng)m>2時(shí),若f'(x)<0,則
-<x<,若f'(x)>0,則
x<-或x>,
所以函數(shù)f(x)的單調(diào)減區(qū)間為
(-,),
函數(shù)f(x)的單調(diào)增區(qū)間為
(-∞,-),(,+∞).
綜上所述,當(dāng)m=0時(shí),函數(shù)f(x)的單調(diào)減區(qū)間為(-∞,0),(0,+∞),無單調(diào)增區(qū)間;
當(dāng)m<0時(shí),函數(shù)f(x)的單調(diào)減區(qū)間為
(-∞,-),(-,),(,+∞),無單調(diào)增區(qū)間;
當(dāng)m>0時(shí),①當(dāng)0<m<2時(shí),函數(shù)f(x)的單調(diào)減區(qū)間為
(-∞,-),(,+∞),函數(shù)f(x)的單調(diào)增區(qū)間為
(-,);
②當(dāng)m=2時(shí),f(x)=0,為常數(shù)函數(shù),無單調(diào)區(qū)間;
③當(dāng)m>2時(shí),函數(shù)f(x)的單調(diào)減區(qū)間為
(-,),函數(shù)f(x)的單調(diào)增區(qū)間為
(-∞,-),(,+∞)…(13分)