设w>0,函数f(x)=2sinwx在【—派/3,派/4】上单调递增,则w的取值范围是?
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设w>0,若函数f[x]=2sinwx在[-π\3,π\4]上单调递增,则w的取值范围是
解析:∵函数f[x]=2sinwx(w>0)在[-π\3,π\4]上单调递增
f(x)单调增区间:wx∈[2kπ-π/2,
2kπ+π/2]==>x∈[2kπ/w-π/(2w),2kπ/w+π/(2w)]
区间[-π/3,π/4]包含于[2kπ/w-π/(2w),2kπ/w+π/(2w)]
∴-π/(2w)<=-π/3==>-1/(2w)<=-1/3==>w<3/2
π/(2w)>=π/4==>1/(2w)>=1/4==>w<=2
取二者交w<=3/2
∴w的取值范围是0<
w<=3/2
解析:∵函数f[x]=2sinwx(w>0)在[-π\3,π\4]上单调递增
f(x)单调增区间:wx∈[2kπ-π/2,
2kπ+π/2]==>x∈[2kπ/w-π/(2w),2kπ/w+π/(2w)]
区间[-π/3,π/4]包含于[2kπ/w-π/(2w),2kπ/w+π/(2w)]
∴-π/(2w)<=-π/3==>-1/(2w)<=-1/3==>w<3/2
π/(2w)>=π/4==>1/(2w)>=1/4==>w<=2
取二者交w<=3/2
∴w的取值范围是0<
w<=3/2
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