已知函数f(x)满足f(x)=f'(1)e^(x-1) - f(0)x+(1/2)x^2 (2)若f(x)≥(1/2)x
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已知函数f(x)满足f(x)=f'(1)e^(x-1) - f(0)x+(1/2)x^2 (2)若f(x)≥(1/2)x^2+ax+b,求(a+1)b的最大值.
1、
f(x)=f′(1)e^(x-1)-f(0)x+1/2x^2中,令x=0的f'(1)=ef(0)
所以f(x)=f(0)e^x-f(0)x+1/2x^2
关于x求导得:f'(x)=f(0)e^x-f(0)+x
故f'(1)=f(0)e-f(0)+1=ef(0)解得f(0)=1
所以f(x)=e^x - x + 1/2 x^2
f'(x)=e^x-1+x
当x>0时,f'(x)>0,函数单调增加
当x
f(x)=f′(1)e^(x-1)-f(0)x+1/2x^2中,令x=0的f'(1)=ef(0)
所以f(x)=f(0)e^x-f(0)x+1/2x^2
关于x求导得:f'(x)=f(0)e^x-f(0)+x
故f'(1)=f(0)e-f(0)+1=ef(0)解得f(0)=1
所以f(x)=e^x - x + 1/2 x^2
f'(x)=e^x-1+x
当x>0时,f'(x)>0,函数单调增加
当x
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