Let f (x) = SiNx cosx + X + 1 (0 < x < 2 π), find the monotone interval and extreme value of function f (x)

Let f (x) = SiNx cosx + X + 1 (0 < x < 2 π), find the monotone interval and extreme value of function f (x)

f(x) =sinx-cosx+x+1
f'(x) = cosx +sinx +1 =0
√2(sin(x+π/4)) = -1
x+π/4 = 5π/4 or 7π/4
x=π or 3π/2
f''(x) = -sinx + cosx
f''(π) = -10 (min)
max f(x) = f(π) = π+2
min f(x)=f(3π/2) = 3π/2
Monotone interval
Increase (0, π] or [3 π / 2,2 π)
Decrease [π, 3 π / 2]

Let f (x) = SiNx / 2 + cosx, find: (1) monotone interval of F (x) (2) If for any x > = 0, there is f (x)

Sin (x / 2) + cosx? Or (SiNx) / 2 + cosx1, (SiNx) / 2 + cosx is assumed to be Cty = 1 / 2, siny = 2 / √ 5F (x) = ctgy * SiNx + cosx = 1 / siny (cosy * SiNx + siny * cosx) = √ 5 / 2Sin (x + arctan2) increasing function: 2K Π - Π / 2 ≤ x + arctan2 ≤ 2K Π + Π / 22K Π - Π / 2-arctan2 ≤ x ≤ 2K

Monotone increasing interval of function f (x) = SiNx + cosx

f(x)=sinx+cosx
=√2sin(x+π/4)
Monotonically increasing interval: [2K π - 3 π / 4,2k π + π / 4] K is an integer

Urgent: find the extremum of the function FX = SiNx (1 + cosx) Why cosx = - 1 is not an extreme point!

F (x) = SiNx (1 + cosx) f '(x) = cosx (1 + cosx) + SiNx (- SiNx) = cosx + cos? X-sin? X = 2cos? X + cosx-1 = (cosx + 1) (2cosx-1) f' (x) = 0, cosx = - 1, cosx = 1 / 2 & n

The known function f (x) = 1 2(sinx+cosx)-1 Then the range of F (x) is () A. [-1，1] B. [- Two 2，1] C. [-1，- Two 2] D. [-1， Two 2]

From question f (x)=
cosx，(sinx≥cosx)
sinx，(sinx＜cosx) =
cosx，x∈[2kπ+π
4，2kπ+5π
4]
sinx，x∈(2kπ-3π
4，2kπ+π
4) ，
When x ∈ [2K π + π
4，2kπ+5π
4] When f (x) ∈ [- 1,
Two
2]
When x ∈ (2k π - 3 π)
4，2kπ+π
4) When f (x) ∈ (- 1,
Two
2）
Therefore, the value range is [- 1,
Two
2]．
Therefore, D

Find the maximum and minimum of F (x) = 3sinx + 2cosx First floor. Why can we do this? Please explain the reasons,

(2); (2) = (2) = (2); (2) = (2); (2) = (2); (2) = (2) = (2); (2) = (2) = (2); (2) = (2); (2) = (2) = (2); (2) = (2); (2) = (2); (2
Because the maximum and minimum values of sin (a + x) are + 1, - 1;
Therefore, the maximum and minimum values of F (x) are root 13 and - radical 13

TaNx = 3 calculates 5cosx + 3sinx

Seven fifths root ten~
TaNx = 3, if the small right angle side is 1, then the other right angle side is 3, and the hypotenuse is the root sign ten~

(4tanx-2cosx) / (5cosx + 3sinx) = 6 / 11 to find TaNx,

Should it start with 4sinx
If so
That is, the numerator denominator is divided by cosx at the same time to get (4tanx-2) / (5 + 3tanx) = 6 / 11, thus TaNx = 2

The function y = x + 2cosx is in the interval [0, π 2] The maximum value on is () A. Two B. π 2+ Three C. π 6+ Three D. π 6+ Two

Y ′ = 1-2sinx = 0, x = π
6 or x = 5 π
6，
So y = x + 2cosx is in the interval [0, π
6] It's an increasing function over the interval [π]
6，π
2] The minus function is up,
Then x = π
At 6, y = π
6+ 
3，x=π
When 2, y = π
2＜π
6+
3，
So the maximum is π
6+
3，
Therefore, C

What is the maximum value of the function y = x + 2cosx on the interval [0, U / 2]?

y'=1-2sinx=0
sinx=1/2
x=π/6
In [0, π / 2], SiNx is an increasing function
So y 'is a minus function
So 0