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)

Analysis: if f (x) = sinx-cosx + X + 1, then f '(x) = cosx + SiNx + 1 = √ 2 (2 / 2cosx + 2 / 2 / 2 SiNx) + 1 = 1 = 2 (sin π / 4 cosx + cos π π / 4sinx) + 1 = 1 = 2 = 2 (x ＋ π / 4) ＋ 1. Let 0 ＜ x ＋ π / 4 ＜ 1. Let 0 ＜ x ＋ π / 4 ＜ π / 2 and 3 π / 2 ＜ 2 π, get - π / 4 ＜ 4 ＜ 4 ＜ 4 ＜ 4 ＜ 4 ＜ 4 ＜ 4 ＜ 4 ＜ 4 ＜ 4 ＜ 4 x ＜ π / 4 and 5 π / 4

Let f (x) = SiNx / (2 + cosx) find the monotone interval of F (x) Let f (x) = SiNx / (2 + cosx) (1) find the monotone interval of F (x) (2) if there is f (x) ax for any x > 0, find the value range of A

(1)f'(x)=(2cosx+1)/(2+cosx)^2
F '(x) > 0, increasing interval, cosx > - 1 / 2, (- 2pi / 3,2pi / 3) + 2K * PI
f'(x)

Who can help me find the extremum of the function f (x) = SiNx + cosx on 0-2 π

When the derivative of F (x) = cosx SiNx = 0, there is an extreme value in the domain of definition
That is, when x = π / 4,5 π / 4, there is an extreme value,
The maximum value is f (π / 4) = radical 2,
The minimum value is f (5 π / 4) = - radical 2
PS: classmate, to find the extremum is to get the derivative of the function and make it 0

Given that f (x) = 2cosx-3sinx, when f (x) is the maximum, TaNx is=

y=2cosx-3sinx=√13cos(x+s)
When cos (x + s) = 1, y has a maximum
x+s=360n
Then TaNx = - tans = - 3 / 2

Given TaNx = 3, calculate the value of (4sinx-2cosx) / (5cosx + 3sinx)

If TaNx = SiNx / cosx = 3, cosx and SiNx are not zero,
So (4sinx-2cosx) / (5cosx + 3sinx)
=(4sinx/cosx-2)/(5+3sinx/cosx)
=(4tanx-2)/(5+3tanx)
=(4*3-2)/(5+3*3)
=10/14
=5/7

Find the maximum value of the root sign (1-cos2x) of the function y = 2cosx + 3, and find the size of TaNx when the function takes the maximum value

The maximum value of the function y = 2cosx + 3 root number (1-cos2x) of the function y = 2cosx + 3 root number (1-cos2x) and the maximum value of the maximum value of the function, the size of TaNx size f (x) = 2cosx + 3 √ (1-cos2x) = 2cosx + 3 √ [(cosx) ^ 2 + (SiNx) ^ 2 - (cosx) ^ 2 + (cosx) ^ 2 + (SiNx) ^ 2] = 2cosx + 3 √ [2 (SiNx) ^ 2] = 2cosx + 3 √ 2 | 2 | SiNx | the maximum value of the maximum value of the maximum value of the maximum value of the maximum value of the maximum value of the maximum value is | 2 ^ 2 ^ 2 + 3 and

The maximum value of the function y = x + 2cosx in the interval [0,1 / 2] ..................

The derivative of the function y = x + 2cosx shows that it is an increasing function at [0,1 / 2], so the maximum value is 1 / 2 + 2cos1 / 2

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

By y = x + 2cosx,
Derivative of X: y '= 1-2sinx,
Let y ′ = 0, i.e. 1-2sinx = 0,
ν SiNx = 1 / 2, when x ∈ [0, π / 2],
x=π/6.
Y max = f (π / 6) = π / 6 + 2 ×√ 3 / 2 = π / 6 + √ 3

What is the maximum value of the function y = x + 2cosx in the interval [0, π / 2], and the approximate solution process

The first derivative of X has y '= 1-2sinx
When x = π / 6, y '= 0
When x ∈ (0, π / 6), y '> 0
x∈（π/6,π/2）y'<0
So when x = π / 6, y = π / 6 + √ 3 is the maximum

What is the value of X when the function y = x + 2cosx gets the maximum value on [0, π / 2] How to get the derivative is y = x + 2cosx y'=1-2sinx Can we get the result by other methods instead of derivation

This is based on the formula that the coefficient is constant
The derivative of y = x ^ n is y '= n × x ^ n-1
The derivative of y = cosx is y '= - SiNx
Remember to use it like this. Check the formula in the textbook