Try to find out an interval of length 1, in which the function X-1 / 3x + 2 has at least one zero operation problem It seems that the substitution of x = 1 is not correct, as if it does not satisfy f (a) f (b) of finding zero

Try to find out an interval of length 1, in which the function X-1 / 3x + 2 has at least one zero operation problem It seems that the substitution of x = 1 is not correct, as if it does not satisfy f (a) f (b) of finding zero

X = 1, the function X-1 / 3x + 2 has a zero point
（1/2,3/2）

Given that f (x) = LG x, any x 1, x 2 (x 1 ≠ x 2) in the domain of definition of function f (x), we have the following conclusion
①0＜f ' (3)＜f(3)-f(2) ＜f ' (2);
②0＜f ' (3)＜f ' (2)＜f(3)-f(2);
③[f(x1)-f(x2)]/(x1-x2)＞0；
④f((x1+x2)/2)＜[f(x1)+f(x2)]/2
Please judge right and wrong first and prove them one by one,

One pair, f '(3) f' (2) tangent slope, the difference is secant slope
2 wrong
Three pairs, increasing function
Error 4. The definition of convex function should be greater than

It is known that the function f (x) = ax ^ 3 + BX ^ 2 + CX passes through the point (- 4 / 3, - 4 / 27), and the maximum value o is obtained when x = - 1
1. Find the analytic expression of F (x)
2. In the interval [M, O] (M

1.
Because f (x) = ax ^ 3 + BX ^ 2 + CX passes through the point (- 4 / 3, - 4 / 27), and the maximum value o is obtained when x = - 1
f’(x）=3ax^2+2bx
So we substitute the point (- 4 / 3, - 4 / 27) into f (x) = ax ^ 3 + BX ^ 2 + Cx
-64/27a+16/9b-4/3c=-4/27
3a-2b=0
-a+b-c=0
The solution is a = 10, B = 15, C = 5
F(X)=10x^3+15x^2+5x
2. F (x) is defined as [M, 0]
f’(x）=30x^2+30x
Let f '(x) = 0, then x = 0 or x = - 1
If f '(x) > 0, then X

Given the function y = AX3 + bx2, when x = 1, there is a maximum of 3; then 2A + B=______ ．

Because the function y = AX3 + bx2, y ′ = 3ax2 + 2bx, and when x = 1, y ′| x = 1 = 3A + 2B = 0, and Y | x = 1 = a + B = 3, that is, 3A + 2B = 0A + B = 3, a = - 6, B = 9, х 2A + B = - 3

Given the function y = ax ^ 3 + BX ^ 2 and x = 1, y has a maximum value of 3, (1) find the value of A.B. (2) increase interval of function y

The derivative of the function is
y'=3ax²+2bx
Let y '= 0
Then 3ax & # 178; + 2bx = 0
The solution is x = - 2b / 3A or x = 0
therefore
-2b/3a=1
2b=-3a ①
When x = 1,
y=a+b=3 ②
From the above two formulas, we can get the following results
a=-6,b=9
therefore
y=-6x³+9x²
The increasing interval is (0,1)

Let f (x) = ax ∧ 3 + BX ∧ 2 + CX obtain the minimum value of - 8 at x = x0, and the image of its derivative function y = f '(x) passes through the points (- 2,0), (2 / 3,0)
(2) If f (x) ≥ m ∧ 2-14m holds for X ∈ [- 3, 3], the value range of real number m is obtained?
Function opening down = = the original function is f (x) = - x Λ 3-2x Λ 2 + 4x

f'(x)=3ax²+2bx+c
so
12a-4b+c=0
4a/3+4b/3+c=0
Solution
b=2a,c=-4a
f'(x)=3ax²+4ax+-4a=a(x+2)(2x-3)
f(x)=ax^3+2ax²-4ax
f(-2)=-8a+8a+8a=8a=-8
a=-1
f(x)=-x^3-2x²+4x
X∈[-3 ,3]
f(-3)=27-18-12=-3
f(3)=-27-18+12=-33
f(2/3)=40/27
In [- 3,3], f (x) min = - 33, f (x) max = 40 / 27
For X ∈ [- 3, 3], f (x) ≥ m ∧ 2-14m is constant
m²-14m≤-33
m²-14m+33≤0
3≤m≤11

Let y = ax ^ 3 + BX ^ 2 + CX + D (a)

Now y '= 3ax ^ 2 + 2bx
From y '= 0
Get X1 = 0 (known and minimum point)
x2= -2b/3a
Therefore, the original function takes the maximum value at x = - 2b / 3a
By substituting x = - 2b / 3A into the original function, we can get the following results
y= (4b^3) / (27a^2)
Let k = - A, then a ^ 2 = k ^ 2, B = 1 + K
y=4(1+k)^3 / (27k^2)
=4/27 * (3+k+3/k+1/k^2)
The formula in brackets is always greater than 0, and tends to positive infinity when k tends to 0 and positive infinity, so the minimum value is the minimum value, so the minimum value of the formula in brackets is required
First and second derivation of the formula in brackets
After a derivation, 1-3 / K ^ 2-2 / K ^ 3 = 0
k^3-3k-2=0
(k+1)^2 * (k-2)=0
K1 = - 1 (rounding, because k = - A is greater than 0) K2 = 2
Quadratic derivation, y '' = 6 / K ^ 3 + 6 / K ^ 4
Substituting k = 2, we get y '' > 0
Therefore, the maximum value of the original function is the minimum when k = 2, when a = - k = - 2, B = 1 + k = 3, the maximum value is 1

When x is equal to 1, a cubic function has a maximum value of 4. When x is equal to 3, it has a minimum value of 0, and the function passes through the origin

y=ax^3+bx^2+cx+d
Because it passes through the origin, the constant term is d = 0
y'=3ax^2+2bx+c
Because the function has a maximum of 4 when x = 1 and a minimum of 0 when x = 3,
So 3ax ^ 2 + 2bx + C = 0 has two real roots 1 and 3
a=1/3,b=-2,c=3
So y = x ^ 3 / 3-2x ^ 2 + 3x

When x = 1, a cubic function has a maximum value of 4, when x = 3, a minimum value of 0, and the function passes through the origin, find the analytic formula of the function and list it,

Let the cubic function equation be y = ax ^ 3 + BX ^ 2 + CX + D, substituting x = 1, y = 4, x = 3, y = 0, x = 0, y = 0 into the equation, then d = 0, then a + B + C = 4, 27a + 9b + 3C = 0. Then the derivative of this function is y '= 3ax ^ 2 + 2bx + C. when x = 1, there is a maximum value, that is, y' = 0, the equation 3A + 2B + C = 0 is obtained. Solving the cubic equation, a = - 1, B = 6, C = - 9

If we know that f (x) is a cubic function, if x = 1, there is a maximum of 4, if x = 3, there is a minimum, and the image of the function crosses the origin, then the analytic expression of the function is

f'(x)=(x-1)(x-3)=x^2-4x+3
The original function is f (x) = (1 / 3) x ^ 3-2x ^ 2 + 3x