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

Updated on amusement 2024-08-05
12 answers
  1. Anonymous users2024-02-15

    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

  2. Anonymous users2024-02-14

    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).

  3. Anonymous users2024-02-13

    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

  4. Anonymous users2024-02-12

    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.

  5. Anonymous users2024-02-11

    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

  6. Anonymous users2024-02-10

    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

  7. Anonymous users2024-02-09

    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).

  8. Anonymous users2024-02-08

    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).

  9. Anonymous users2024-02-07

    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 .

    The math tutoring team will answer the questions for you, quality assurance].

    If my is helpful to you, please choose it as a satisfactory answer in time, thank you.

  10. Anonymous users2024-02-06

    An informal solution is used here, but theoretically there should be nothing wrong.

  11. Anonymous users2024-02-05

    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=

  12. Anonymous users2024-02-04

    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

Related questions
14 answers2024-08-05

1。Do it down. Bring in the endpoint value.

2。Discriminant equations for quadratic equations. >>>More

12 answers2024-08-05

f'(x)=3x^2+2bx+c

So g(x)=x 3+(b-3)x 2+(c-2b)x-cg(-x)=-x 3+(b-3)x 2-(c-2b)x-c is an odd function. g(-x)=-g(x) >>>More

14 answers2024-08-05

1) f(x)=ax 3+bx 2-2x+c, then f'(x)=3ax^2+2bx-2

and f'The two roots of (x)=0 are x=-2 and x=1 >>>More

11 answers2024-08-05

Solution: a=3 to the 55th power = (3 to the fifth power) to the 11th power. >>>More

8 answers2024-08-05

f(x)=x 2+ax-a+3=(x+a2) -a 4-a+3, i.e., f(x) is a parabola with an open phase and an axis of symmetry x=-a2. >>>More