f(X)为偶函数在区间(0，正无穷）上是增函数,若x属于[1/2,1],不等式f(ax+1)≤f(x-2)恒成立，a的取值范围 f(x)是偶函数，且f(x)在(0,正无穷)上是增函数，若x...
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这是一个含参不等式在区间上恒成立的问题 需要求参数范围
是一种常见的题型。
一楼二楼的 都是错误的解法
我做出来的答案是 开区间下 -2到-1
考虑f(x)在区间[0，正无穷）上是增函数时,可得在(负无穷，0]上是减函数.
f(ax+1)≤f(x-2)恒成立，说明，|ax+1|≤|x-2|，因为x属于[1/2,1],所以x-2<0
即|ax+1|≤2-x，
当ax+1≥0时
ax+1≤2-x
=> (a+1)x≤1
=> a≤1/x-1 (当x属于[1/2,1])
=> a≤0
ax+1≤0
-1-ax≤2-x
=> a≥1-3/x (当x属于[1/2,1])
=> a≥-2
所以-2≤a≤0
因为函数的递增区间不包括0，而题中函数也没有强调f（0）是否有意义，就必须刨除，即使有意义也可能在0点是跳跃函数，那么必须计算ax+1≠0的情况，这时当x属于[1/2,1]时，解得a<-2或a>-1，
于是最终结果为-1＜a≤0
如下：
闭区间下-2,0
详细解答 请看 连接
我以高考数学130分的成绩 向你保证 解法 是完全正确的
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