1.设g(x)=ax^+bx+c,
g(1)=a+b+c=-1,①
g(x-1)+g(1-x)
=a(x-1)^+b(x-1)+c
+a(1-x)^+b(1-x)+c
=2ax^-4ax+2a+2c
=x^-2x-1,
比较系数得2a=1,2a+2c=-1,
解得a=1/2,c=-1,
代入①,b=-1/2.
∴g(x)=(1/2)x^-(1/2)x-1.
2.存在x>0,使得f(x)=g(x+1/2)+mlnx+9/8=(1/2)x^+mlnx>=0,
x=1时上述不等式恒成立,
∴m的取值范围是R.
3.H(x)=f(x)-(m-1)x=(1/2)x^-(m-1)x+mlnx,x∈[m,1],m>0,
H'(x)=x-(m-1)+m/x
=[x^-(m-1)x+m]/x
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