(1) f'(x)=e^x-e^(-x)
f'(x)=0 x=0
f(-1)=1/e+e≈ 3.0862 f(0)=2 f(ln2)=2+1/2=2.5
∴ max(f(x))=1/e+e≈ 3.0862
(2) a=-1 f(x)=e^x+e^(-x) g(x)=2x-x^2
e^x-e^(-x)=2-2x x0=0.490073
(3) f(x)-g'(x)=e^x+e^(-x)-2-2ax
[f(x)-g'(x)]'=e^x-e^(-x)-2a=0 (e^x)^2-2ae^x-1=0 x=a+√(1+a^2) x=a-√(1+a^2)(舍去)
[f(x)-g'(x)]'‘=e^x+e^(-x)>0
即x=a+√(1+a^2)时,f(x)-g'(x)取得最小值,要使不等式f(x)≥g‘(x)恒成立,必须且只需:
e^(a+√(1+a^2) )+e^(-(a+√(1+a^2) ))-2-2a(a+√(1+a^2) )>=0
==>a