题目
A.f(2)<f(3)<g(0) | B.g(0)<f(3)<f(2) | C.f(2)<g(0)<f(3) | D.g(0)<f(2)<f(3) |
答案
又∵f(x)-g(x)=ex
∴解得:f(x)=
ex-e-x |
2 |
ex+e-x |
2 |
故f(x)单调递增,又f(0)=0,g(0)=-1,有g(0)<f(2)<f(3)
故选D.
A.f(2)<f(3)<g(0) | B.g(0)<f(3)<f(2) | C.f(2)<g(0)<f(3) | D.g(0)<f(2)<f(3) |
ex-e-x |
2 |
ex+e-x |
2 |