Question

f (x) =根号2sin (x +π/4),x属于【-π/4,2π/3】求值域

Answer 1

x属于【-π/4,2π/3】
x
+π/4属于【0,11π/12】
当x=-π/4时sin
(x
+π/4)取最小值0，2sin
(x
+π/4)取最小值0
当x=π/4时sin
(x
+π/4)取最大值1，2sin
(x
+π/4)取最大值2
根号2sin
(x
+π/4)值域是【0，根号2】

Answer 2

f(x)=2sin^2(π/4-x)-
√
3cos2x
=1-cos(π/2-2x)-
√
3cos2x
=1-sin2x-
√
3cos2x
=1-2(1/2sin2x+
√
3/2cos2x)
=1-2(sin2xcosπ/3+
sinπ/3cos2x)
=1-2sin(2x+π/3)
故最小正周期t=π
当2kπ+π/2≤2x+π/3≤2kπ+3π/2时，函数单调递减
即函数的单调递减区间为：kπ+π/12≤x≤kπ+7π/12
当0＜x＜π/2时
π/3＜2x+π/3＜4π/3
故当2x+π/3=π/2时
即x=π/12
f(x)min=f(π/12)=-1