Find the value range of F (x) = - cos ^ 2x - SiNx + 3

Find the value range of F (x) = - cos ^ 2x - SiNx + 3

f(x)=-cos^2x-sinx+3
=-(1-sin^2x)-sinx+3
=sin^2x-sinx+2
=(sinx-1/2)^2+7/4
∵sinx∈[-1,1]
∴(sinx-1/2)^2∈[0,9/4]
∴(sinx-1/2)^2+7/4∈[7/4,4]
The range is [7 / 4,4]

Given the function y = 2Sin (x / 2 + Pai / 3), find the maximum value of function and the set of X corresponding to the maximum value Finding monotone increasing interval of function

The maximum value of 1 is obtained when x ∈ {x 124x = π / 2 + 2 + 2n π, n ∈ Z} when x ∈ {x ә x ∈ {x | x = π / 2 + 2 + 2n π, n ∈ Z, n ∈ Z, so we can get the maximum value of Y, the maximum value of Y is 2, when x ∈ {x | x | x = π / 3 + 3 + 4N π / 3 + 4N π, n ∈ Z}, monotonic increasing interval is (- 5 π / 3 + 3 + 4N π, π / 3 + 3 + 4N π), n ∈ Z ∈ Z, Z, Z, Z, n it's a good idea~

I am a freshman in senior high school. Although I have been studying for more than a month, I feel confused about the function in mathematics. My junior high school foundation is not good. When I was in high school, I felt that the teacher could not keep up with the homework. In fact, I worked hard, but I couldn't keep up with the teacher's progress. What's the way? Teach me!

Senior one should take time, the problem of difficult function content is very difficult, but the basic should be firmly grasped, no matter in the future selection of articles or theories, the mathematical foundation should be solid
Senior one function has nothing to do with junior high school. Don't think about how the foundation of junior high school is not good. First of all, we should grasp the present
As long as you can prove the monotonicity of a function, you will find that it is meaningless to do and do only those problems: the parity of function is not difficult, so take a good look in the book and master the definition proof; there are several comprehensive problems of function, some of which are particularly difficult, and the key is analysis. It is easy to write. You should be calm, otherwise you will not be able to get to senior two or senior three in the future, Senior one is mainly a foundation. It doesn't matter if you don't do well in the exam. As long as you understand the wrong topic, it will be of great help to you in the future
Don't worry! It's normal that I can't keep up with you, but I usually have to be a step faster than others. For example, it's better to preview before class, and usually do some extracurricular exercises

Let f (x) = 2Sin (2x - π / 6) 2cos2x. Find the minimum of function f (x) 1... Known function f (x) = 2Sin (2x - π / 6) 2cos2x. Find the minimum positive period of function f (x)... 2... Known vector a = (1,3) vector b = (4, - 2) find the absolute value of 2a-b

F (x) = 2 (sin2xcos π / 6-cos2xsin π / 6-cos2xsin π / 6) + 2cos2x = sin2x-cos2x + 2cos2x = sin2x + 2cos2x = sin2x + cos2x = 2 (/ 2sin2x + 1 / 2cos2x) = 2 (COS π / 6sin2x + sin π / 6cos2x) = 2Sin (2x + π / 6), so the maximum value of F (x) is 2. Cycle T = π 2, 2a-b = (- 2,8) | 2a-b | = = (2,8) | 2a-b | = = = (2,8) | (2a-b) | = (- 2,8) | 2a-b √ (- 2) ^ 2 +

Given that f (x) is a quadratic function, f (0) = 1, f (x + 1) = f (x) + X + 1, find the analytic formula of function f (x)

Let f (x) = ax ^ 2 + BX + CF (0) = C = 1F (x + 1) = a (x + 1) ^ 2 + B (x + 1) + C = ax ^ 2 + (2a + b) x + A + C + BF (x) + x + 1 = ax ^ 2 + (B + 1) x + A + C + BF (x) + X + 1 = ax ^ 2 + (B + 1) x + C + 1A x ^ 2 + (2a + b) x + A + C + B = ax ^ 2 + (B + 1) x + C + 1, therefore, 2A + B = B + 1, a + C + B = C + 1, and C = 1, a = 1 / 2, B, B = 1 / 2, B = 1 / 2, B is a = 1 / 2, B is a = 1 / 2, B = c= 1 / 2, C = 1F (x) = 1 / 2x ^ 2 + 1 / 2x + 1

Function problem f (x) = follow sign 3 (sin3x / 2 * cosx / 2 + cos3x / 2 * SiNx / 2)+ F (x) = the following sign 3 (sin3x / 2 * cosx / 2 + cos3x / 2 * SiNx / 2) + cos ^ 2 x-sin ^ 2 x (1) find the minimum positive period of function f (x); (2) find the monotone increasing interval of function f (x) Cos square x sin square x

f(x)=√3sin(3x/2+x/2) +cos2x
= √3sin2x +cos2x
=2sin(2x+π/6)
So the minimum positive period is t = 2 π / 2 = π
Let 2K π - π / 2 ≤ 2x + π / 6 ≤ 2K π + π / 2
K π - π / 3 ≤ x ≤ K π + π / 6 is obtained
The increasing interval is [K π - π / 3, K π + π / 6], K ∈ Z

Let f (x) = m (cosx + SiNx) ^ 2 + 1-2sin ^ 2 x, X belongs to R and y = f (x) Through point (PI / 4,2) 1, find the value of real number m, find the minimum value of function f (x) and the set of X at this time

If (x) = m (sin ^ 2x + cos ^ 2x) + msin2x + cos2x = m + msi2x + cos2x = m + msi2x + cos2xf (π / 4) = 2 = m + m, (sin π / 2 = 1, cos π / 2 = 0) M = 1F (x) = 1 + sin2x + cos2x = 1 + √ 2Sin (2x + π / 4) minimum value is 1 - √ 2, at this time, 2x + π / 4 = 2K π / 4 = 2K π + π 2x 2x = 2K π + 3 π / 4x = 2K π + 3 π / 4x = K π / 4x = k π + 3 π / 4x = k π = k π + 3 π / 4x = + 3 π / 8, K ∈ Z

If the analytic expression of the function f (x) is f (x) = 2tanx - (2Sin ^ 2 (x / 2) - 1) / (SiNx / 2 * cosx / 2) 30 - resolution time: 2007-11-3 12:34 1. If the analytic expression of the function f (x) is f (x) = 2tanx - (2Sin ^ 2 (x / 2) - 1) / (SiNx / 2 * cosx / 2) Then f (π / 12) is 2. Known that α is an acute angle and sin α = 4 / 5 (1) find the value of sin ^ 2 + sin2 α / cos ^ 2 α + Cos2 α (answer 20) (2) Tan (α - 5 π / 4) (answer 1 / 7)

1) F (x) = 2tanx - (2Sin ^ 2 (x / 2) - 1) / (SiNx / 2 * cosx / 2) f (x) = 2tanx + cos X / (1 / 2 * SiN x) = 2tanx + 2 Cotx = 2 [(SiN x / cos x) + (cos X / SiN x)] = 4 / sin (2x) bring in π / 12, = 8 1) find sin ^ 2 + sin2 α / cos ^ 2 α + Cos2 α sin α = 4 /

(COS (1 / x) square + 1) x power X tends to 0, find the limit

1 < = (COS (1 / x) square + 1) x power < = 2 ^ x
The left and right sides tend to 1 at the same time
The limit is 1

It is known that the minimum positive period of the function y = 1-2sin (ω X - π / 3) (ω > 0) is 4 (1) Finding the maximum value of a function and the value set of the independent variable x when taking the maximum value (2) finding the monotone interval of the function

T=4
w=π/2
So y = 1-2sin (π X / 2 - π / 3)
So π X / 2 - π / 3 = 2K π - π / 2,
X = 4k-1 / 3, the maximum value is 1 + 2 = 3
Similarly, x = 4k-5 / 3, the minimum value is 1-2 = - 1
As y increases, sin (π X / 2 - π / 3) decreases
So 2K π + π / 2