Question

已知函数f(x)=x^2／x+1,（1）若f(x)在点（1，f(1)）处的切线斜率为1／2，求实数a的值，（2）若f(x)在x=1处

已知函数f(x)=x^2+a／x+1,（（2）若f(x)在x=1处 取得极值，求函数f(x)的单调区间

Answer 1

(1)
f(x)=(x^2+a)／(x+1)
f'(x) = [(x^2+a)- 2(x+1)x ]/ (x+1)^2
f'(1) = (1+a -4)/4 = 1/2
-3+a = 2
a = 5
(2)
f'(1) = 0
=> (1+a -4)/4 = 0
a = 3
f'(x) = [(x^2+3)- 2(x+1)x ]/ (x+1)^2
= [-x^2-2x+3]/(x+1)^2
put f'(x) = 0
-x^2-2x+3 =0
-(x+3)(x-1) = 0
x = 1 or -3

f''(x) =[(-x^2-2x+3)2(x+1) - (x+1)^2(-2x-2)]/(x+1)^4
f''(1) = [0- 4(-4)]/16 = 1 >0 (min)
f''(-3) <0 (max)
increasing (-∞, -3]∪[1,+∞)
decreasing [-3,1]

Answer 2

题目未详！