求 y=e的x次方-x²+2ax(a＞ln2-1)的单调区间

求 y=e的x次方-x²+2ax(a＞ln2-1)的单调区间
求 y=e的x次方-x²+2ax(a＞ln2-1)的单调区间

求 y=e的x次方-x²+2ax(a＞ln2-1)的单调区间
f(x)=e^x-x²+2ax
f'(x)=e^x-2x+2a.
f''(x)=e^x-2.
x∈(-∞,ln2)时,f''(x)＜0,x∈(ln2,+∞)时,f''(x)＞0.
故f'(x)在(-∞,ln2)递减,(ln2,+∞)递增.
因此,f'(x)min=f'(ln2)=2-2ln2+2a=2(a+1-ln2).
∵a＞ln2-1
∴a+1-ln2＞0.
故f'(x)min＞0,即f'(x)＞0恒成立.
∴f(x)在R上递增.
综上,f(x)的单调区间为R,且在R上单调递增.