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Solution: The necessary conditions for a polynomial function to take the minimum are: the first derivative = 0, and the second derivative (the derivative of the first derivative) is greater than zero.
Derivative: f'(x) = 3x^2 - 3b = 3(x^2 - b),f''(x) = 6x。
Order f'(x) = 0, resulting in x 2 = b. We're going to talk about it here.
1) If b < 0, then x has no solution, there is no extreme point (of course, there is no minima), so b must "=0;".
2) If b = 0, then x = 0, but at this point the second derivative f''(0) = 0, this point is not an extreme point (in fact, it is an inflection point), so b=0 is not allowed;
3) If b >0, there are two possible extreme points: x1 = sqrt(b), x2 = -sqrt(b), substitute them into f''(x), discovery.
f''(x1) >0, which meets the conditions of the minimum point, so x1 is the minimum point. As long as b is greater than zero, there is a solution of x1, so in the end, all the allowable values of b are b>0
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f'(x)=3x^2-3b
f'(x)=3x 2-3b has zero point at (0,1), b>0,3x 2-3b=0 ==> x=- b, x= b needs 0< b<1==> 00 to meet the title.
Therefore, the range of values of b that satisfies the condition is (0,1).
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f Tease (x) = 3x 2-3b
When b 0 is off-topic.
When b 0 f(x)=0.
x= b or x=- b (rounding).
The cubic of the function f[x]=x -3bx+3b has a minimum value in [0,1].
Full front finger height 0 b wisdom ruler 1
The function f[x] = the cubic of x - 3bx + 3b
0<b<1
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f(x)=x 3-3bx+3b, then:
f'(x)=3x 2-3b, let f'(x)=3x 2-3b=0, gets: x 2=b, the function f(x)=x 3-3bx+3b has a minimum in (0,1), so b>0, so x=- b, or x= b, x<- b, f'(x)>0;
b b, f'(x)>0。
So the function f(x) obtains a minimum value at x= b, so 0< b<1, so 0 is the range of values of b.
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f′(x)=3x^2-3b
When b 0, it is not spring, and the sky is in line with the topic.
When b 0 f(x)=0.
x= b or x=- b (rounding).
The three-fold circle blind power -3bx+3b of the function f[x]=x has a minima cavity in [0,1].
Meet 0 b 1
The function f[x] = the cubic of x - 3bx + 3b
0<b<1
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Answer: 0 "Lead Nazhi = b< = 1
f'(x)=3x 2-3b =0, the solution obtains, x=- root number b, or x=root number b, there is a small value of extreme sensitivity spring in [0,1], so Huai eliminate 0<=b<=1
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From the meaning of the title, f (x) = 3x2-3b, so that f (x) = 0, then x=
b and the function f(x)=x3-3bx+b has a minimum value in the interval (0,1), 0 b
1, b (0,1), so the answer is (0,1).
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Solution: From the problem, the derivative of the function f(x)=x3-6bx+3b is f (x)=3x2-6b has zero points in (0,1), and f (0) is 0, f (1) 0 is -6b 0, and (3-6b) 0
0 b 1 dress up the tomb 2, so the answer is: (0, 1 2).
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Analysis: f'(x)=3x²-3b
Because the function f(x)=x 3bx+3b has a minimum value in (0,1), it is discussed in two cases.
3b<0, 3-3b>0
Solution: 0 b 3
3b>0,3-3b<0
The solution set of b is an empty set.
Therefore, the value range of b is .
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An informal solution is used here, but theoretically there should be nothing wrong.
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