2010浙江高考数学 选择题 求解过程,网上答案不够详细

2024-12-23 07:12:50
推荐回答(5个)
回答1:

把f(x)看成是p(x)=4sin(2x+1)和q(x)=-x的迭加.划出两个简图,你会发现当x=-4时fx)>0,x=-2时,f(x)>0,在-4
同理可分析其他情况.由于用作图法较简单,所以是数形结合

回答2:

用计算器table功能

回答3:

画y=x与y=4sin(2x+1) 图像。。看交点是哪个区间

回答4:

这样的,f(x)=0时,函数f(x)=4sin(2x+1)-x=0,函数g(x)=4sin(2x+1)和u(x)=-x
分别画两个函数的图像。g(x)=4sin(2x+1)的图像有sinx图像经过平移转换得到的。(这叫转化思想)两个图像的交点就是0点。看看落在哪个区间就行了。(画图和计算结合就叫数形结合。)

回答5:

f(x)的零点就是h(x)=4sin(2x+1)和g(x)=x两个函数图象的交点,在同一坐标系中作出两个函数图象,很容易发现h(x)=4sin(2x+1)在(-4,-2)大于0,而g(x)=x在(-4,-2)小于0,两函数图象没有交点,所以函数f(x)=4sin(2x+1)-x在(-4,-2)不存在零点

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