Question

已知函数f(x)=2sin(π-x)cosx 求f(x)在区间[-π/3,π/6]上的最大值和最小值

需要过程的

Answer 1

f(x)=2sin(π-x)cosx =2sinxcosx=sin2x
由正弦函数图像可知：
x在区间（-π/3，-π/4）内单调递减
在区间（-π/4，π/6）内单调递增
所以f(x)在x=π/6处取最大值 f(x)max=f(π/6)=sin2*π/6=2分之根号3
f(x)在x=-π/4处取最小值 f(x)min=f(-π/4）=sin(-π/2)=-1

Answer 2

f(x)=2sin(π-x)cosx =2sinxcosx=sin2x
f(x)在区间[-π/3,-π/4]上递减；在区间[-π/4,π/6]上递增

所以f(x)在x=π/6处取最大值 f(π/6)=sin2*π/6=2分之根号3
f(x)在x=-π/4处取最小值 f(-π/4）=sin(-π/2)=-1

Answer 3

f(x)=sin2x
最大值为sin2*π/6=√3/2
最小值为-1