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Let f (x) = ax ^ 3-3x + 1 (x belongs to R). If f (x) is greater than or equal to 0 for any x belonging to (0,1), then the value range of real number a is
For any x belonging to (0,1), f (x) is greater than or equal to 0
That is, ax & # 179; ≥ 3x-1, a ≥ 3 / X & # 178; - 1 / X & # 179; always holds
Let g (x) = 3 / X & # 178; - 1 / X & # 179;, 00, G (x) increase progressively
1/2
Given f (x) = (AX + 3) ^ 2, prove f (1), at least one of F (2) is less than or equal to 1
Assume that all are greater than 1
f(1)=(a+3)^2>1
a+31
a-2
f(2)=(2a+3)^2>=1
2a+3=1
a-1
So the title is obviously wrong
For example, when a = 1
f(1)=(1+3)^2>1
f(2)=(2+3)^2>1
All greater than 1
Given that f (x) = x ^ 2 + ax + 3-A, if x belongs to [- 2,2], f (x) is greater than or equal to 0, then the range of a is obtained
There is a possibility
1. Discriminant = a ^ 2-12 + 4A = 0
3) Symmetry line - A / 2 = 2
The intersection of the above three conditions is - 7
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