已知函数f(t)=t^0.5+1⼀t^0.5+(t+1⼀t+1)^0.5,函数g(t)=t^0.5+1⼀t^0.5-(t+1⼀t+1)^0.5,求f(t)的最小值和g(t)

如题
2024-12-19 13:49:22
推荐回答(1个)
回答1:

f(t)有最小值,g(t)有最大值:
f(t)=√t+1/√t+√(t+1/t+1),定义域:t>0
√t+1/√t≥2
t+1/t+1≥3
故:f(t)≥2+√3,当t=1时,等号成立
故f(t)的最小值:2+√3
g(t)=√t+1/√t-√(t+1/t+1),定义域:t>0
g(t)=√(t+1/t+2)-√(t+1/t+1)
=1/[√(t+1/t+2)+√(t+1/t+1)]
=1/[t+1/√t+√(t+1/t+1)]
=1/f(t)
f(t)≥2+√3,故:0故g(t)的最大值:2-√3