高一数学求值域的方法及详解 这几个我知道,换元法,配凑法,反表示和分离常数不用说,其他的能经常用的

2025-03-16 19:00:51
推荐回答(2个)
回答1:

除楼主说的几种方法,还有很多,比如:

(1)构造法(包括构造向量法,构造复数法,构造图形法等)
[举例]求f(x)=√(2-x)+√(x-1)的值域.
构造向量m=(1,1),n=(√(2-x),√(x-1)),
则依向量模不等式|m|·|n|≥|m·n|得
(1²+1²)[(2-x)+(x-1)]≥[1·√(2-x)+1·√(x-1)]²
→0<√(2-x)+√(x-1)≤√2.
故所求函数值域为(0,√2].

(2)判别式法
[举例]求y=x²+4x+9值域.
原式即x²+4x+(9-y)=0.
上式判别式不小于0,即
16-4(9-y)≥0
解得,y≥5.
即函数值域为[5,+∞).

(3)不等式法:
[举例]在定义域(0,π/2)内,求三角函数f(x)=1/sinx+8/cosx值域.
依权方和不等式得
f(x)=1/sinx+8/cosx
=1^(3/2)/(sin²x)^(1/2)+4^(3/2)/(cos²x)^(1/2)
≥(1+4)^(3/2)/(sin²x+cos²x)^(1/2)
=5√5,
故所求函数值域为[5√5,+∞).

(4)求导数法
[举例]x>0时,求f(x)=(㏑x)/x的值域.
原式求导得f′(x)=(1-㏑x)/x².
f′(x)>0时,0f′(x)=0时,x=e;
f′(x)<0时,x>e.
∴f(x)|max=f(e)=1/e.
故函数值域为(-∞,1/e]

(5)线性规划法
[举例](略)

(6)函数单调性法
[举例](略)

还有很多方法,不一一列举了!

回答2:

图像法
函数的单调性法
基本不等式法

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