If the function f x x3 to the power of 3bx 3b has a minimum value at 0,1, find the range of the valu

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

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).

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

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.

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

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

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).

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).

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.

f'(x)=4x^3+3ax^2+4x=x(4x^2+3ax+4),f"(x)=12x^2+6ax+4, f"(0)=4>0, so f(0) is the minimum.

Only x=0 is the extreme point, then the equation 4x 2+3ax+4=0 has no real root or equal root, i.e., delta=9a 2-4*4*4<=0, gets:

8/3=

f'(x)=4x 3+3ax 2+4x, so that it is equal to 0 i.e. x=0, or 4x 2+3ax+4=0

Because the function f(x) only has an extreme value at x=0, 4x 2+3ax+4=0 has no solution.

i.e. (3a) 2-64<0

Solution -8 3