Is f (x) = cos (x + π) even function

Is f (x) = cos (x + π) even function

solution
f(x)=cos(x+π)
=-cosx
∵f(-x)=-cos(-x)=-cosx=f(x)
That is even function

Given that f (x) is an odd function defined on the interval [- 1,1] and is an increasing function, f (x) = 1 (1) solves the inequality f (x + 1 / 2)

Because f (x) is an odd function defined on the interval [- 1,1]
Then - 1 ≤ (x + 1 / 2) ≤ 1, the solution is - 3 / 2 ≤ x ≤ 1 / 2 (1)
-1 ≤ (1-x) ≤ 1, the solution is 0 ≤ x ≤ 2 (2)
According to (1) and (2), 0 ≤ x ≤ 1 / 2 (3)
Because f (x) is an increasing function, and f (x + 1 / 2)

The function y = f (x) (x ≠ 0) is an odd function and an increasing function when x ∈ (0, + ∞). If f (1) = 0, the inequality f [x (x-1 / 2)]

f(-1)=f(1)=0,f[x(x-1/2)]

Given the function f (x), f (1) = 1, and f '(x)

Let g (x) = f (x) - X / 2-1 / 2, then the solution set of the original inequality is equivalent to: G (x)

Given the function f (x) = (5 ^ x-1) / (5 ^ x + 1), X ∈ [- 1,1]; (1) solve the inequality f (x)

f(x)=(5^x-1)/(5^x+1)
=(5^x+1-2)/(5^x+1)
=1-2/(5^x+1)
∵ 5 ^ x is an increasing function
5 ^ x + 1 is an increasing function
2 / (5 ^ x + 1) is a decreasing function
1-2 / (5 ^ x + 1) is an increasing function
F (x) is an increasing function
f(x)

Given the function f (x) = | x-a |, where a > 1 (1) when a = 2, find the solution set of the inequality f (x) ≥ 4 - | x-4 |; (2) given the solution set {x | 1 ≤ x ≤ 2} of the inequality | f (2X + a) - 2F (x) | ≤ 2 about X, find the value of A

(1) When a = 2, f (x) ≥ 4 - | x-4 | can be reduced to | X-2 | + | x-4 | ≥ 4; when x ≤ 2, we get - 2x + 6 ≥ 4, and the solution is x ≤ 1; when 2 < x < 4, we get 2 ≥ 4, and there is no solution; when x ≥ 4, we get 2x-6 ≥ 4, and the solution is x ≥ 5; so the solution set of the inequality is {x | x ≥ 5 or X ≤ 1}. (2) let H (x) = f (2x + a) - 2F

Given the vector a = (2cos2x, sinxcosx), B = (a, b), f (x) = vector a × vector B - √ 3 / 2, the image of function f (x) is about the line x = π / 12
And f (0) = √ 3 / 2
(1) Finding the minimum positive period and monotone increasing interval of F (x)
(2) How to translate the image of function into even function
On the symmetry of the line x = π / 12

F (x) = 2acos (2x) + bsinxcosx - √ 3 / 2 = 2acos (2x) + B / 2 sin (2x) - √ 3 / 2F (0) = 2A - √ 3 / 2 = √ 3 / 2A = √ 3 / 2F (x) = √ 3 cos (2x) + B / 2 sin (2x) - √ 3 / 2 = √ (3 + B & # 178 / 4) sin (2x + φ) - √ 3 / 2, where Tan φ

Find the minimum value of the function y = 2cos2x + 5sinx-4

Y = 2cos2x + 5sinx-4 = 2 (1-sin2x) + 5sinx-4 = - 2sin2x + 5sinx-2. Let SiNx = t (- 1 ≤ t ≤ 1). The original function is y = - 2t2 + 5t-2. The equation of axis of symmetry is t = 54 ＞ 1. Y = - 2t2 + 5t-2 is an increasing function on [- 1, 1]. Ymax = - 2 × 12 + 5 × 1-2 = 1, ymax = - 2 × (- 1) 2 + 5 ×

Given the function y = ACOS (2x + π 3) + 3, the maximum value of X ∈ [0, π 2] is 4, then the value of real number a is______ ．

When a ＞ 0, - a ≤ ACOS (2x + π 3) ≤ 12a, ∵ ymax = 4, ∵ 12a + 3 = 4, ∵ a = 2; when a ＜ 0, 12a ≤ ACOS (2x + π 3) ≤ - A, similarly, 3-A = 4, ∵ a = - 1

Given the function y = ACOS (2x + Pie / 3) - 3, X belongs to [0, Pie / 2], the maximum value is 4, find the value of real number a

acos（2x＋π/3）－3≤4
acos（2x＋π/3）≤7
x∈〔0,π/2〕2x∈[0,π] 2x+π/3∈[π/3,π4/3]
When x = 0, cos (2x + π / 3) has a maximum, cos π / 3 = 1 / 2
In this case, a / 2 = 7
a=14