It is known that the functions y = ax and y = - BX are decreasing functions in the interval (0, + ∞). Try to determine the monotone interval of the function y = AX3 + bx2 + 5

It is known that the functions y = ax and y = - BX are decreasing functions in the interval (0, + ∞). Try to determine the monotone interval of the function y = AX3 + bx2 + 5

∵ functions y = ax and y = - BX are both decreasing functions in the interval (0, + ∞), a < 0, B < 0. From y = AX3 + bx2 + 5, we get y ′ = 3ax2 + 2bx. Let y ′ > 0, that is, 3ax2 + 2bx > 0, ■ - 2b3a < x < 0. Therefore, when x ∈ (- 2b3a, 0), the function is an increasing function; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; Let y ′ < 0, i.e. 3ax2 + 2bx < 0, X ＜ - 2b3a or x > 0. Therefore, when x ∈ (- ∞, - 2b3a) and (0, + ∞), the function is a decreasing function; the monotone increasing interval of the function y = AX3 + bx2 + 5 is (- 2b3a, 0); the monotone decreasing interval is (- ∞, - 2b3a) and (0, + ∞)

It is known that the function f (x) = ax3-x2 + BX + 2 (a, B ∈ R) is an increasing function in the interval (- ∞, 0) and (4, + ∞), and a decreasing function in the interval (0, 4). (I) find the value of a and B; (II) find the tangent equation of the curve y = f (x) at x = 1

(1) ∵ f '(x) = 3ax2-2x + B, and f (x) is an increasing function in the interval (- ∞, 0) and (4, + ∞), and a decreasing function in the interval (0, 4), ∵ f' (0) = 0, B = 0. And f '(4) = 0, a = 16. (II) ∵ f (x) = 16x3 − x2 + 2, we get f' (x) = 12x2 − 2x. When x = 1, f '(1) = − 32. At this time, y = f (1) = 76. That is, the slope of the tangent is - 32, and the tangent coordinates are (1, 76) The tangent equation is 9x + 6y-16 = 0

The function f (x) = ax + B / 1 + X2 is odd in the domain (- 1.1), and f (1 / 2) = 2 / 5
1. Find the analytic expression of the definite function f (x)
2. Prove that f (x) is an increasing function on (- 1.1) by using the domain of definition
3. Solve the inequality f (t-1) + F (T) less than 0

1.f(0)=0+b=0,b=0
f(x)=ax,f(1/2)=a/2=2/5,a=4/5
f(x)=4x/5
2. Ren-1

It is known that the function f (x) = ax + bx2 + 1 is an odd function defined on (− 1,1), and f (12) = 25. (1) determine the analytic expression of function f (x); (2) judge the monotonicity of function f (x) and prove it when x ∈ (- 1,1); (3) solve the inequality f (2x-1) + F (x) < 0

(1) From the meaning of the question, we can see that f (- x) = - f (x) ∪ ax + bx2 + 1 = - ax + bx2 + 1 ∪ ax + B = - ax-b, ∪ B = 0 ∫ f (12) = 25, ∪ a = 1 ∪ f (x) = XX2 + 1; (2) when x ∈ (- 1,1), the function f (x) increases monotonically. It is proved that: F ′ (x) = (1 − x) (1 + x) (x2 + 1) 2, X ∈ (- 1,1) ∪

It is known that the function f (x) = (x2 + ax + b) / X is an odd function over the domain {XLX ∈ R, X ≠ 0}
(1) When b > 0 and x > 0, we prove that f (x) > F (b) (3) if f (1) = 2, we can find the range of function f (x)

The function f (x) = (X & # 178; + ax + b) / (x) is an odd function, then: a = 0. In this case, f (x) = (X & # 178; + b) / (x), because f (1) = 2, then: B = 1, then: F (x) = (X & # 178; + 1) / (x) first proves that f (x) decreases on (0,1) and increases on (1, + ∞). Then when x > 0, the range of F (x) is [2, + ∞). Considering that the function is

It is known that the square of 1 + X over F (x) = ax + B is an odd function defined on (- 1,1), and f (1 / 2) = 2 / 5
1. Find the analytic expression of function f (x)
2. Prove that f (x) is an increasing function on (- 1,1) by definition

1. The odd function of F (x) on (- 1,1)
∴f(0)=b=0
And ∵ f (1 / 2) = (A / 2 + b) / (1 + 1 / 4) = 2 / 5
∴a=1
∴f(x)=x/(1+x^2）
(2) Set - 1

The function f (x) = the square of X + ax + B is an odd function defined on R, and f (& frac12;) = two fifths
(1) Find the values of real numbers a and B, and determine the analytic expression of F (x)
(2) It is proved by definition that f (x) is an increasing function in the interval (- 1,1)

∵ f (x) = (AX + b) / (X & # 178; + 1) is an odd function defined on R
∴f(-x)=-f(x)
∴(-ax+b)/(x²+1)=-(ax+b)/(x²+1)
∴-ax+b=-ax-b
∴b=0
∴f(x)=ax/(x²+1)
∵f(1/2)=2/5
∴(a/2)/(1/4+1)=2/5
∴a=1
∴f(x)=x/(x²+1)
(2)
Set - 1

F (x) = ax + B1 + X2 is a function defined on (- 1,1), whose image passes through the origin, and f (12) = 25. (1) determine the analytic expression of the function f (x); (2) prove by definition that f (x) is an increasing function on (- 1,1)

(1) F (12) = 25.12, and f (12) = 25.12, f (12) = 25.12, f (12) = 25.6 \ (12) = 25 (12) = 25 \ (12) = 25 \12 = 12 \\\\12 = 25 \\ = 25 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\0; ■ (x1 −) X2) (1 − x1x2) (1 + X21) (1 + X22) < 0f (x1) - f (x2) < 0  f (x1) < f (x2) 〉 f (x) is an increasing function on (- 1,1);

The function f (x) = (AX + b) / (the square of 1 + x) is an odd function defined on (- 1,1), and f (1 / 2) = 2 / 5
(1) Find the analytic expression of function f (x) (2) prove that f (x) is an increasing function on (- 1,1) by definition

Because: F (1 / 2) = 2 / 5
So: A / 2 + B = 1 / 2
Because: F (x) is an odd function
So f (- x) = - f (x), that is (- ax + b) / (1 + x2) = - (AX + b) / (1 + x2),
That is (- ax + b) = - (AX + b)
That is, B = - B
So B = 0, a = 1
（1）f(x)=x/(1+x2)
(2) Set - 1

Let a function have a second derivative on [0,1], and | f '' (x) | ≤ m, and f (x) has the maximum value in [0,1]. It is proved that | f (0) | + | f (1) | ≤ M
High number
Let a function have a second derivative on [0,1], and | f '' (x) | ≤ m, and f (x) has the maximum value in (0,1). It is proved that | f (0) | + | f (1) | ≤ M
It's parentheses

How can this be possible
f(x)=-x^2-10,M=2
f"(x)=-2,
The maximum value in [0,1] is - 10,
And | f (0) | + | f (1) | = 10 + 11 = 21 > M
It's not good to use parentheses, such as f (x) = - (x-0.5) ^ 2-10
f"(x)=-2,
The maximum value is - 10 in (0,1)
|f(0)|+|f(1)|=21>M