Question

已知函数f(x)=1/2x^2-alnx(a属于R) 1.若函数f(x)的图像在x=2处的切线方程为y=x+b...

已知函数f(x)=1/2x^2-alnx(a属于R)

1.若函数f(x)的图像在x=2处的切线方程为y=x+b,求a,b的值
2.若函数f(x)在[1,正无穷)为增函数,求a的取值范围

Answer 1

1.f'(x)=x-a/x,
f'(2)=2-a/2=1,a=2.
∴f(x)=x^2/2-2lnx,f(2)=2-2ln2,
切线方程为y-(2-2ln2)=x-2,即y=x-2ln2,b=-2ln2.

2.x>1时f'(x)=(x^2-a)/x>0,
∴a<=1.

Answer 2

f'(x) = x - a/x
f'(2) = 1, so 2-a/2 = 1, a = 2
f(2) = 2 - 2ln2
切线方程 y = x + b = 2-2ln2, so b = -2ln2

2. if f'(x) > 0 when x >= 1
x-a/x >= 0
x^2 - a >=0
a <= x^2
so a <= 1