Question

设g(x)=e^x,f(x)=g[λx+(1-λ)a-λg(x)],其中a,λ是常数,且0<λ<1.求函数f(x)的极值

Answer 1

λx+(1-λ)a-λg(x)
=λx+(1-λ)a-λe^x

f(x)
=g[λx+(1-λ)a-λg(x)]
=e^[λx+(1-λ)a-λe^x] 1

g'[λx+(1-λ)a-λg(x)]
={e^[λx+(1-λ)a-λe^x]}'
=e^[λx+(1-λ)a-λe^x] * [λx+(1-λ)a-λe^x]'
=e^[λx+(1-λ)a-λe^x] *(λ-λe^x)
=λ(1-e^x)e^[λx+(1-λ)a-λe^x]=0
∵e^[λx+(1-λ)a-λe^x]>0
∴1-e^x=0
x=0
把x=0代入 1式得
f(0)
=e^[(1-λ)a-λ]
∴极值点是(0,e^[(1-λ)a-λ])

Answer 2

楼主的题目打错了吧。。