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“a≤3”是“函数f(x)=x2-4ax+1在区间[4,+∞)上为增函数”的( )
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高二下学期数学《》真题及答案
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已知函数fx=x3+ax+b的图像是曲线C直线y=kx+1与曲线C相切于点13. 1求函数fx的解
已知a>0且a≠1下列函数中在区间0a上一定是减函数的是
f(x)=
f(x)=a
x
f(x)=log
a
(ax)
f(x)=x
2
﹣3ax+1
a=1是函数fx=x2-4ax+3在区间[2+∞上为增函数的________条件.
设函数fx=x3-1+ax2+4ax+24a其中常数a>1则fx的单调减区间为________.
若函数fx=lgx2+ax﹣a﹣1在区间[2+∞上单调递增则实数a的取值范围是
(﹣3,+∞)
[﹣3,+∞)
(﹣4,+∞)
[﹣4,+∞)
已知函数fx=x3+x2+ax+1.Ⅰ若曲线y=fx在点01处切线的斜率为﹣3求函数fx的单调区间Ⅱ
定义在R.上的函数fx是偶函数且fx=f2-x.若fx在区间[12]上是减函数则fx
在区间[-2,-1]上是增函数,在区间[3,4]上是增函数
在区间[-2,-1]上是增函数,在区间[3,4]上是减函数
在区间[-2,-1]上是减函数,在区间[3,4]上是增函数
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在区间[-2,-1]上是增函数,在区间[3,4]上是增函数
在区间[-2,-1]上是增函数,在区间[3,4]上是减函数
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