Given the function f (x) = x ^ 2 + 10x-a + 3, X ∈ [- 2, + ∞), if f (x) > = 0 holds, find the value range of A

Given the function f (x) = x ^ 2 + 10x-a + 3, X ∈ [- 2, + ∞), if f (x) > = 0 holds, find the value range of A

f(x)=x^2+10x-a+3
=(x+5)^2-a-22
Axis of symmetry x = - 5
F (x) increases monotonically on X ∈ [- 2, + ∞)
If f (x) > = 0 is constant, only the minimum value of F (x) > = 0 is needed
That is, f (- 2) > = 0
4-20-a+3>=0 a

The known function f (x) = 34sin & nbsp; x-14cos & nbsp; X. (1) if cosx = - 513, X ∈ [π 2, π], find the value of function F & nbsp; (x); (2) shift the image of function f (x) to the right M units, so that the translated image is symmetrical about the origin. If 0 < m < π, try to find the value of M

(1) Because cosx = - 513, X ∈ [π 2, π], SiNx = 1213, so, f (x) = 34 × 1213 + 14 × 513 = 3313 + 552 (2) f (x) = 34sinx-14cosx = 12sin (x - π 6). So, the image of F (x) is shifted 5 π 6 units to the right, and the image of y = - 12sinx is obtained

If x ∈ (- ∞, 0), f (x) + XF '(x)

Let g (x) = XF (x), G (x) be an even function, G (0) = 0, G '(x) = f (x) + XF' (x)
When x0, G (x) is monotonically decreasing and increasing, G (x) > = 0
And 0

Let F X be an odd function defined on R, and f (2) = 0. When x > 0, f (x) > XF '(x) holds, then the solution set of inequality X & # 178; f (x) < 0 is

Constructor f (x) = - x ^ 3 + 4x, f (x) is an odd function on R, f (2) = 0,
f'(x)=-3x^2+4,
x> When f (x) - XF '(x) = - x ^ 3 + 4x-x (- 3x ^ 2 + 4) = 2x ^ 3 > 0,
If f (x) is a function satisfying the problem, then
x^2*f(x)=-x^3*(x+2)(x-2)

Given the singular function f (x) defined on R, let its derivative be f '(x). When x ∈ (- ∞, 0], the value range of real number x with constant x f' (x) f (2x-1) is?

∵ f (x) is an odd function
∴×f'(x)

F (x) + XF '(x) when x ∈ (- infinity, 0)

A:
F (x) is an odd function, f (- x) = - f (x)
G (x) = XF (x), the domain of definition is the same as f (x), which is a real number range R and symmetric about the origin
g(-x)=(-x)f(-x)=-x*[-f(x)]=xf(x)=g(x)
So: G (x) is an even function
g'(x)=f(x)+xf'(x)

Why is f (x) = MSIN (AX + b) (a > 0) an increasing function in the interval [a, b], and f (a) = - MF (b) = m function g (x) = MCOs (AX + b) the maximum value m in [a, B]

Look at the picture. Take ax + B as t
On the increasing interval of Sint, there is the maximum value of cost, and the maximum value = M

If 0 ≤ α ≤ π / 2, f (MSIN α) + F (1-m) > 0, the range of M is determined

F (x) is an increasing odd function
From F (MSIN θ) + F (1-m) > 0,
That is, f (MSIN θ) > F (m-1)
∴msinθ＞m-1,∴1＞m(1-sinθ).
When θ = θ / 2, the inequality holds
When 0 ≤ θ＜ / 2, m ＜ 1 / (1-sin θ),
The minimum value of ∵ 1 / (1-sin θ) is 1,
∴m＜1.
one

If the function f (x) = MSIN (ω x + φ) (ω > 0) is an increasing function in the interval [a, b], and f (a) = - m, f (b) = m, then the function g (x) = MCOs (ω x + φ) is ()
A. Is an increasing function B. is a decreasing function C. can obtain the maximum MD. can obtain the minimum - M

∵ function f (x) is an increasing function in the interval [a, b], and f (a) = - m, f (b) = m adopts the special value method: let ω = 1, φ = 0, then f (x) = msinx, let the interval be [- π 2, π 2]. ∵ m > 0, G (x) = mcosx does not have monotonicity in [- π 2, π 2], but has the maximum value m, so choose: C

Given the function f (x) = 2sinx + 1, (1) let the constant ω > 0, if y = f (ω x), is an increasing function in the interval [- π / 2,2 π / 3], the value range of ω is obtained
The answer is that the process is π / 2W ≥ 2 π / 3, - π / 2W ≤ - π / 2,
I don't know. How did π / 2W, - π / 2W come from,

The period of F (ω x) is 2 π / ω, obviously in the interval [- 2 π / ω / 4,2 π / ω / 4] is increasing, so π / 2W ≥ 2 π / 3, - π / 2W ≤ - π / 2 is OK
For example, the period of SiNx is 2 π, increasing at [- 2 π / 4,2 π / 4]