What is the maximum and minimum of the function y = x-sinx on (π / 2, π)?

What is the maximum and minimum of the function y = x-sinx on (π / 2, π)?

Because its derivative is greater than 0, it increases monotonically on R, so the maximum value π and the minimum value π / 2-1

Monotone decreasing interval of sine function

Y = SiNx decreases monotonically in [pi / 2 + 2K * PI, 3pi / 2 + 2K * PI] (k belongs to all integers)
In addition:
If a sine function is a composite function y = sin [u (T)], then you just need to replace the above x with u (T), that is, u (T) belongs to [pi / 2 + 2K * PI, 3pi / 2 + 2K * PI] (k belongs to all integers), and then find out the value of T

The maximum value of function f (x) = XX + 1 is______ ．

According to the meaning of the question, if x ≥ 0, then f (x) = XX + 1 = 1 x + 1 x and X + 1 x ≥ & nbsp; 2, then f (x) ≤ 12, so the answer is 12

Y = sin3x + cos3x what is the value of X when the function takes the maximum and minimum values?

When y = sin3x + cos3x = radical 2 sin ((3x + π / 4) 3x + π / 4 = 2K π + π / 2, x = 2K π / 3 + π / 12, the maximum value is 3x + π / 4 = 2K π - π / 2, and the minimum value is x = 2K π / 3 + π / 4

Find the period, maximum and minimum of the following functions y = sin3x * cos3x
For a certain process, thank you

∵y=sin3x*cos3x
=(1/2)sin6x
The period T = 2 π / 6 = π / 3
∵sin6x∈[-1,1]
∴y=(1/2)sin6x∈[-1/2,1/2]
The maximum value is 1 / 2
The minimum is - 1 / 2

The maximum of function y = sin3x-cos3x

Y = sin3x sin (π / 2-3x) = 2cos [(3x + π / 2-3x) / 2] sin [(3x + π / 2 + 3x) / 2] = 2cos (π / 4) sin (3x + π / 4) = √ 2Sin (3x + π / 4), so the maximum value is √ 2 if and only if 3x + π / 4 = 2K π + (π / 2), (k = 1,2,3 ）Then x = (2 / 3) k π + (π / 12), (k = 1,2,3 )...

We know that vector a = (2cos & # 178; X, √ 3), vector b = (1, sin2x), function f (x)
We know that vector a = (2cos & # 178; X, √ 3), vector b = (1, sin2x), function f (x) = vector a × vector B-1, G (x) = vector B & # 178; - 1
1. Find the solution set of equation g (x) = 0
Finding the minimum period of positive interval function

1 g(x)=b²-1=(1,sin2x)(1,sin2x)-1=1+sin²2x-1=sin²2x
∧ g (x) = 0 = > Sin & # 178; 2x = 0 = > 2x = k π = > x = k π / 2, K is any integer
The solution set of {g (x) = 0 is {K π / 2: K ∈ Z}
2 f(x)=ab-1=(2cos²x,√3)(1,sin2x)-1=2cos²x+√3sin2x-1
=cos2x+√3sin2x=2sin(2x+π/6)
The minimum positive period of F (x) is 2 π / 2 = π, and the monotone increasing interval is
2kπ-π/2≤2x+π/6≤2kπ+π/2 => 2kπ-2π/3≤2x≤2kπ+π/3
=>K π - π / 3 ≤ x ≤ K π + π / 6, namely [K π - π / 3, K π + π / 6]

Let f (x) = 2cos & sup2; X + sin2x + A, find the minimum positive period and monotone increasing interval of (1) function

Minimum positive period π
F (x) = radical 2 * sin (2x + π / 4) + A + 1
2K π - π / 2 with monotone increasing

Given Tan x = - 2 and X is the angle of the second quadrant, we can find SiN x and COS X
fast

(secx)^2=1+(tanx)^2=5
X is the angle of the second quadrant, secx

Y = sinxcosx / (2 + SiNx + cosx) minimum

Hello!
Let u = 2 + SiNx + cosx = 2 + √ 2 sin (x + π / 4) ∈ [2 - √ 2,2 + √ 2]
(u-2)² = sin²x+cos²x+2sinxcosx = 1+2sinxcosx
sinxcosx = [ (u-2)²－1) ] /2
y = [ (u-2)²－1) ] / (2u)
= (u² - 4u+3) / (2u)
= u/2 + 3/(2u) －2
= 1/2 (u + 3/u) － 2
≥ √ (U * 3 / U) - 2 [basic inequality (a + b) / 2 ≥ √ (AB)]
= √3 －2
If and only if u = 3 / u, i.e. u = √ 3, take the equal sign
U = √ 3 in [2 - √ 2,2 + √ 2], the equal sign can be obtained
So the minimum value is √ 3-2