函数f(x)=2^2+a·log2(x^-2)+b,在x=1/2时有最小值1,求a、b

题目

函数f(x)=2^2+a·log2(x^-2)+b,在x=1/2时有最小值1,求a、b

答案

函数f(x)=2[log2(x)]^2+a·log2[x^(-2)]+b,在x=1/2时有最小值1,求a、b.
f'(x)=4*[log2(x)][1/(x ln2)]+a·[-2x^(-3)]/[x^(-2)*ln2]=
=4*[log2(x)]/(x ln2)-2ax^(-3)/[x^(-2)*ln2],
函数有最小值,它必先是极小值,故
f'(x)=4*[log2(x)]/(x ln2)-2ax^(-3)/[x^(-2)*ln2]=0,
2[log2(x)]/(x ln2)=ax^(-3)/[x^(-2)*ln2],
2[log2(x)][x^(-2)*ln2]=ax^(-3)(x ln2),
设这个极小值是最小值,它在x=1/2处,则代入x=1/2:
2[log2(1/2)][(1/2)^(-2)*ln2]=a(1/2)^(-3)((1/2) ln2),
-2[4ln2]=a*4ln2,
a=-2.
函数f(x)=2[log2(x)]^2+a·log2[x^(-2)]+b,在x=1/2时有最小值1,
则 f(1/2)=2[log2(1/2)]^2+a·log2[(1/2)^(-2)]+b=1,
代入a=-2,
2-2*2+b=1,
b=3.