三角函数图像的一道题，当然会追加分数的

显示全部楼层 · 2009-8-21 15:20:36

已知函数f(x)=2sinwx(w>0)在区间[-π/3，π/4]上的最小值是-2，则w的最小值等于多少？设f(t)=2sinwt=-2, t∈[-π/3,π/4] --->wt=(2k+1)π --->t=(2k+1)π/w....∈[-π/3,π/4] (2k+1)π/w≥-π/3--->w≥3(2k+1) (2k+1)π/w≤π/4---->w≥4(2k+1) ∵w＞0,∴w取最小值时k=0--->w=4...

千问 · 2009-8-21 15:20:36

设f(t)=2sinwt=-2, t∈[-π/3,π/4] --->wt=(2k+1)π --->t=(2k+1)π/w....∈[-π/3,π/4] (2k+1)π/w≥-π/3--->w≥3(2k+1) (2k+1)π/w≤π/4---->w≥4(2k+1) ……但X在[-π/3 , π/4]上无需能取到极大值2，即w≥3能满足题...

千问 · 2009-8-21 15:20:36

w≥-3(2k-1/2) k=0w=1.5...