Let f x sin 2x 3 , find f n x and f 9 3 2

y=sin(2x-3)

y'=2cos(2x-3)=2sin[(π/2)+(2x-3)]y''=-2²sin(2x-3)=2²siny'''=-2³cos(2x-3)=2³siny^(4)=2^4sin(2x-3)=2^4siny^(n)=2^n sin

y^(9)*(3/2)

2^9*sin*(-3/2)

3*2^8*sin(15)

768*sin15

If you don't understand, you can ask, if it helps, please choose satisfied!

I guess I should find the nth derivative of f(x).

f(x)=sin(2x-3)

f'(x)=cos(2x-3)*2

f''(x)=-sin(2x-3)*2^2f'''(x)=-cos(2x-3)*2^3f'''''(x)=sin(2x-3)*2^4f^(4k)*(x)=sin(2x-3)*2^4kf^(4k+1)*(x)=cos(2x-3)*2^(4k+1)f^(4k+2)*(x)=-sin(2x-3)*2^(4k+2)f^(4k+3)*(x)=-cos(2x-3)*2^(4k+3)f^(9)*(3/2)=f^(4*2+1)*(3/2)cos(2x-3)*2^9 (x=-3/2)cos(-3-3)*2^9

2^9*cos6

f(1) =sqrt(2)/2

f(2) =1

f(3) =sqrt(2)/2

f(4) =0

f(5) =sqrt(2)/2

f(6) =1

f(7) =sqrt(2)/2

f(8) =0

Circulate. The sum of the 8 rings of each belt is 0

2014 divided by 8 = 251 remaining rows of bends 6

f(1)+f(2)+f(3)+.f(2014)=f(1) =sqrt(2) 2+1+sqrt(2) 2+0-sqrt(2) 2-1=sqrt(2) 2=root number 2

Summary. Dear I am glad to answer for you: 1) Minimum positive period: 2 Monotonic decreasing interval: [ 3, 2 3] (2) Value range: [-1 2, 1 2].

11.It is known that f(x)=1 2sin(2x+ 3)) 2x+3), x r

1) Find the minimum positive period and monotonic decreasing interval of f(x);

2) Find the range of f(x) over the interval [-4), 4)].

11.The process f(x)=1 2sin(2x+ 3)) 2x+3) is known. 1) Find the minimum positive period and monotonic decreasing interval of f(x); , x∈r11.

It is known that f(x)=1 2sin(2x+ 3)) 2x+3)2) finds the range of f(x) over the interval [-4), 4)].

1) Find the minimum positive period and monotonic decreasing interval of f(x); , x∈r11.It is known that f(x)=1 2sin(2x+ 3)) 2x+3)2) finds the range of f(x) over the interval [-4), 4)].

1) Find the minimum positive period and monotonic decreasing interval of f(x); , x∈r11.It is known that f(x)=1 2sin(2x+ 3)) 2x+3)2) finds the range of f(x) over the interval [-4), 4)].

1) Find the minimum positive period and monotonic decreasing interval of f(x); , x∈r11.It is known that f(x)=1 2sin(2x+ 3)) 2x+3).

f`(x)=6x^2+3

sin** didn't write Lu Sue defeated.

If it is sin (number) it is a number, and the derivative of premature fibrillation is 0sina) = cosa

sin7a) = 7cos7a (composite function).

[[1]]

Actually, the problem is to determine the parity of the function f(x).

First, define the domain symmetry with respect to the origin.

Set the question to be constant: 2f(-sinx)+3f(sinx)=sin2x Replace the above x with -x, you can get.

2f(sinx)+3f(-sinx)=-sin2x.

f(sinx)+f(-sinx)=0.

Combined with the question: 2f(-sinx)+3f(sinx)=sin2x, f(sinx)=sin2x=2sinxcosx, that is, f(sinx)=2sinxcosx

Let k=sinx, easy to know, cosx= (1-k )and -1 k 1 f(k) = 2k (1-k )1 k 1 Obviously, this function is odd.

This is not a contradiction.

f(sinx)=-f(-sinx)

Explain that the function about sinx is an odd function.

Therefore this function is also an odd function.

It mainly uses currency exchange.

Let t=sinx

There is f'(t)=1+arcsint

Get f'(x)=1+arcsinx

The integral f(x)=x+xarcsinx+ (1-x 2)+cc is an arbitrary constant on both sides.