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
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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
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
(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)
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)
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