Question

求下列函數(shù)的值域和最值:

(1)y=2sinx-1;

(2)y=3sin(3x+)+2;

(3)y=2cos²x+5sinx-4;

(4)y=.

Answer 1

思路分析:利用|sinx|≤1,通過變量代換轉(zhuǎn)化為基本函數(shù).

解：(1)∵-1≤sinx≤1,

∴-2≤2sinx≤2.故-3≤2sinx-1≤1.

當(dāng)x=2kπ+(k∈Z)時(shí),y有最大值1;

當(dāng)x=2kπ-(k∈Z)時(shí),y有最小值-3.值域?yàn)椋?3,1］.

(2)u=3x+,則有y=3sinu+2,

∴值域?yàn)椋?1,5］.

當(dāng)u=2kπ+(k∈Z),即x=kπ+(k∈Z)時(shí),y有最大值5.

當(dāng)u=2kπ-(k∈Z),即x=kπ-(k∈Z)時(shí),y有最小值-1.

(3)設(shè)sinx=u,則|u|≤1,y=2cos²x+5sinx-4=2-2sin²x+5sinx-4=-2u²+5u-2.①

問題轉(zhuǎn)化為在定義域［-1,1］內(nèi)求二次函數(shù)①的值域問題.配方,有y=-2(u-)²+,

∵-1≤u≤1,

∴當(dāng)u=-1,即x=2kπ-(k∈Z)時(shí),y有最小值-9;當(dāng)u=1,即x=2kπ+(k∈Z)時(shí),y有最大值1.

∴函數(shù)y的值域?yàn)椋?9,1］.

(4)原函數(shù)可化為y=,即y=1-.

∵1≤sinx+2≤3,

∴≤≤1,

1≤≤3,-3≤≤-1.

故-2≤1≤0.

∴函數(shù)y的值域?yàn)椋?2,0］,并且當(dāng)x=2kπ+時(shí),y=0;當(dāng)x=2kπ-時(shí),y=-2.