Question

已知函数f(x)=(x-1)/(x+a)+ln(x+1),其中实数a不等于1 (1)若a=-2,求曲线y=f(x)在点(0,f(0))处的切线方程

(2)若f(x)在x=1处取得极值，试讨论f(x)的单调性

Answer 1

当a=-2时f(x)=(x-1)/(x-2)+ln(x+1)
所以f＇(x)=(x-2-x+1)/(x-2)^2+1/(x+1)=-1/(x+2)^2+1/(x+1)
f＇(0)=-1/4+1=3/4
所以切线斜率k=3/4
f(0)=1/2+ln1=1/2
所以切线为y-1/2=3/4x
即:y=3/4x+1/2

Answer 2

a=-2
f(x)=(x-1)/(x-2)+ln(x+1)
f'(x)=((x-2)-(x-1))/(x-2)^2+1/(x+1)
=-(x-2)^-2+(x+1)^-1
f'(0)=-1/4+1=3/4
f(0)=-1/a=1/2
切点为（0,1/2）
则切线方程为：y=3/4x+1/2

剩下的写不下了。

Answer 3

（1）f（x）=(x-1)/(x-2) +ln(x+1), f'(x)=-1/(x-2)(x-2) +1/(x+1)，所以f(0)=1/2 , 切线斜率k= f ' (0)=3/4，切线方程为3x-4y+2=0
（2）（-1,1）和（7，+无穷）单调增，（-无穷，-1）和（1,3）和（3,7）递减

Answer 4

当a=-2时f(x)=(x-1)/(x-2) ln(x 1)
所以f＇(x)=(x-2-x 1)/(x-2)^2 1/(x 1)=-1/(x 2)^2 1/(x 1)
f＇(0)=-1/4 1=3/4
所以切线斜率k=3/4
f(0)=1/2 ln1=1/2
所以切线为y-1/2=3/4x
即:y=3/4x 1/2
参考资料：第二问请看: zhidao.baidu.com/question/316995391.html