(x ^ 2-1) SiNx + xcosx

(x ^ 2-1) SiNx + xcosx

=2xsinx+(x²-1)cosx+cosx-xsinx
=xsinx+x²cosx

Monotone decreasing interval of F (x) = cosx SiNx

f(x)=√2(√2/2cosx-√2/2sinx)
f(x)=√2cos(x+π/4)
2kπ

Monotone decreasing interval of y = SiNx + cosx (x ∈ [0,2 π])

The original function is: y = root 2 * sin (x + Π / 4)
From 2K Π + Π / 2

Any x belongs to R, cos2x + SiNx + A is greater than or equal to 0

cos2x+sinx+a>=0
-1

limx→0 (e^x-1)x^2/x-sinx

limx→0 (e^x-1)x^2/【x-sinx】
Law of Robita
=limx→0 【2x(e^x-1)+e^x*x^2】/（-cosx）
=0

It is proved that the minimum positive period of the function f (x) = SiNx (x belongs to R) is 2 Π

Let t and f (x) = the minimum positive period of SiNx
Firstly, it is proved that t = 2 π satisfies the following conditions
f(x+2π)=sin(x+2π)
=sinx
=f(x)
Then, it is proved that t = 2 π is minimum
Anyway, if t ∈ (0,2 π) is periodic
Let x = π / 2
sin(T+π/2)=sin(π/2)=1
In (0,2 π), t + π / 2 ∈ (π / 2,5 π / 2), sin (T + π / 2) ≠ 1
T ∈ (0,2 π) does not hold
In conclusion, the minimum positive period is 2 π

Let f (x) = | x-a | + 2x, where a > 0.1, when a = 2, find the solution set of the inequality f (x) > = 2x + 1

FX ≥ 2x + 1 | X-2 | + 2x ≥ 2x + 1 | X-2 | ≥ 1 x ≥ 3 or - 1 ≥ x (2) always has FX > 0 | x-a | + 2x > 0 when x ≥ A: 3x-a > 03x > a because x ∈ (- 2, + infinity), so, - 6 > a conforms to X ≥ a, so, - 6 > a holds when a > X: x + a > 0A ＞ - x because x ∈ (- 2

Let even function FX = 2X-4, then the solution set of inequality f (X-2) > 0

Do even functions f (x) have defined fields

If the function y = f (x) is even on R and x > = 0, f (x) = x ^ 2-2x-3, then the inequality f (x-1)

x> When f = 0, f (x) = x ^ 2-2x-3,
F (x) = f (- x) = (- x) ^ 2 - 2 (- x) - 3 = x ^ 2 + 2x-3
x

We know that vector a = (with 3 SiNx, cosx) vector b = (cosx, cosx) f (x) = 2 vector a times vector B + 2m-1, x, m belong to R
(1) Find the expression of F (x) with respect to x, and find the minimum positive period
(2) If x belongs to (0,2 / π), the minimum value of F (x) is 5 and the value of M is obtained
(0,2 π) is a closed interval, but I can't find that bracket

1. Vector a * vector b = (√ 3 * SiNx) * cosx + cosx * cosx = √ 3 / 2 * sin2x + 1 / 2 * cos2x + 1 / 2 = sin (2x + Π / 6) + 1 / 2, f (x) = 2, vector a multiplied by vector b + 2m-1, x, m belongs to r = 2 * [sin (2x + Π / 6) + 1 / 2] + 2m-1 = 2Sin (2x + Π / 6) + 2m. T = 2 Π / 2 = Π. 2. X belongs to [0