求y=tan2xtan3x的值域,π⼀4小于x小于π⼀2

2025-06-20 16:41:06
推荐回答(1个)
回答1:

y=tan2xtan^3x
y=2tanx^4/(1-tanx^2)
=(2tanx^4-2+2)/(1-tanx^2)
=[2(tanx^2-1)(tanx^2+1)]/(1-tanx^2)+2/(1-tanx^2)
=-2(tanx^2+1)+2/(1-tanx^2)
=-2(tanx^2-1)+2/(1-tanx^2)-4
=-{[2(tanx^2-1)]+[2/(tanx^2-1)]}-4
π/41
上式≤-2{√[2(tanx^2-1)]*[2/(tanx^2-1)]}-4=-8
当且仅当2(tanx^2-1)=2/(tanx^2-1)时,即tanx=根号2时取等号
最大值为-8