来自董双梅的问题
设函数f(x)=1-e-x.(Ⅰ)证明:当x>-1时,f(x)≥xx+1;(Ⅱ)设当x≥0时,f(x)≤xax+1,求a的取值范围.
设函数f(x)=1-e-x.
(Ⅰ)证明:当x>-1时,f(x)≥xx+1;
(Ⅱ)设当x≥0时,f(x)≤xax+1,求a的取值范围.
1回答
2020-03-22 11:20
设函数f(x)=1-e-x.(Ⅰ)证明:当x>-1时,f(x)≥xx+1;(Ⅱ)设当x≥0时,f(x)≤xax+1,求a的取值范围.
设函数f(x)=1-e-x.
(Ⅰ)证明:当x>-1时,f(x)≥xx+1;
(Ⅱ)设当x≥0时,f(x)≤xax+1,求a的取值范围.
(1)当x>-1时,f(x)≥xx+1当且仅当ex≥1+x令g(x)=ex-x-1,则g'(x)=ex-1当x≥0时g'(x)≥0,g(x)在[0,+∞)是增函数当x≤0时g'(x)≤0,g(x)在(-∞,0]是减函数于是g(x)在x=0处达到最小值,因而当x...