f x 1 x 2 1 x 2 4 1 x 4 8 1 x 8 1 x 8 16 1 x 16 and find the value of f 2 .

You write 1 1+x as (1-x) (1-x 2) 1 (1-x 2)-x (1-x 2) 2 3+1 (1-x 2).

f(x)=2 3+1 (1-x 2)+2 1+x 2+4 1+x 4+8 1+x 8+16 1+x 16 (using squared difference).

2/3+1/（1+x^2）+2/1-x^4+4/1+x^4+8/1+x^8+16/1+x^16

2/3+1/（1+x^2）+2/1+x^4+4/1-x^8+8/1+x^8+16/1+x^16

2/3+1/（1+x^2）+2/1+x^4+4/1+x^8+8/1-x^16+16/1+x^16

2/3+1/（1+x^2）+2/1+x^4+4/1+x^8+8/1+x^16+16/1-x^32

2/3+[f(x)-1/3］/2+16/1-x^32

Finishing: f(x) 2 1 2+16 1-x 32

f(x)＝1+32/（1-x^32）＝x^32-33/（x^32-1）

When x 2:

f（2）＝1294967263/4294967295

The method is as follows, please comma circle for reference:

If there is help from the landslide, please celebrate.

Summary. f(x +2)=x (x +4) find f(x)limx(x -x+a x-2) 3 find ax 2lim after the x don't want to play more. Good.

Summary. Hello, dear, glad to answer for you, f(1 x-x)=2x +2 x =2(x +1 x)=2(1 x-x) +4 and then change the yuan, replace 1 x-x with x, so f(x)=2x +4

f(1 x-x)=2x +2 x, find what f(x) is equal to.

Hello, dear, I'm glad to answer for you, f(1 x-x)=2x +2 x =2(simple x +1 x)=2(1 x-x) +4 and then block the change of yuan, the residual dust is replaced by x 1 x-x, so f(x)=2x +4

Hello, there are no specific steps.

I'm a little bit confused, can you write a complete step and send it to me.

Kiss,**Inside is the problem-solving process.,Take a look at it.。

What I don't understand.

The main thing is to match the back part with 1 x-x by square, why is it added to the fourth.

Because it is necessary to cancel out the -4 in front.

Summary. Good.

f(x)=-2x³+3x²+6

Good. The first defines the set of real numbers in the domain r

I want the process. The first one is not a process.

OK the third.

It says little brother. Good.

The method is as follows, please comma circle for reference:

If there is help from the landslide, please celebrate.

5.Since e(ix)=cos[x]+isin[x], the result is the real part of the integral e ((2+i)x)dx.

re[∫e^((2+i)x)dx]

re[e^((2+i)x)

2+i)] 2cos[x]+sin[x])e (2x) 5, substitution value (e pi-2) 5

10.∫sin[x]^2/x^2dx

∫sin[x]^2

d(1/x)=∫

1/xd(sin[x]^2)

sin[x]^2/x=∫

sin[2x]/x

dx-sin[x]^2/x

Since sin[x]x has an integral value of pi 2 on [0, infinity], the result is pi 2

g(x)=1-x+x²/2-x³/3+……x^2013/2013

f(x)+3=0 or g(x)-3=0

h(x)=f(x)+3=4+x-x²/2+x³/3-……x^2013/2013

h'(x)=1-x+x^2-..x^2012

x = -1, h'(1)=2013>0

x = -1, h'(x)=1-x+x^2-..x^2012=(-x)^2013-1]/[(-x)-1]=(x^2013+1)/(x+1)

x>-1,h'(x)=(x^2013+1)/(x+1)>0

x<-1,h'(x)=(x^2013+1)/(x+1)>0

h'(x)>0 is constant, and h(x) is the increasing function.

h(0)=4

h(-1)=3+1-1-1/2-1/3-1/4-..1/2013

3∴h(-1)<0

f(x)+3=0 only 1 real number solution belongs to (-1,0).

i(x)=g(x)-3

Equally ego i'(x)=-1+x-x^2+..x^2012 <0

i(x) is a subtraction function.

i(0)=-2<0

i(-1)=-3+(1+1+1/2+1/3+..1/2013)>0

g(x)-3=0 and only 1 solution belongs to (-1,0).

The real numbers of f(x)=0 are all in the interval (-1,0).

The minimum value of b-a is 1

g(x)=1-x+x²/2-x³/3+……x^2013/2013

f(x)+3=0 or g(x)-3=0

h(x)=f(x)+3=4+x-x²/2+x³/3-……x^2013/2013

h'(x)=1-x+x^2-..x^2012

x = -1, h'(1)=2013>0

x = -1, h'(x)=1-x+x^2-..x^2012=(-x)^2013-1]/[(-x)-1]=(x^2013+1)/(x+1)

x>-1,h'(x)=(x^2013+1)/(x+1)>0

x<-1,h'(x)=(x^2013+1)/(x+1)>0

h'(x)>0 is constant, and h(x) is the increasing function.

h(0)=4

h(-1)=3+1-1-1/2-1/3-1/4-..1/2013

3∴h(-1)<0

f(x)+3=0 only 1 real number solution belongs to (-1,0).

i(x)=g(x)-3

Equally ego i'(x)=-1+x-x^2+..x^2012 <0

i(x) is a subtraction function.

i(0)=-2<0

i(-1)=-3+(1+1+1/2+1/3+..1/2013)>0

g(x)-3=0 and only 1 solution belongs to (-1,0).

The real numbers of f(x)=0 are all in the interval (-1,0).

The minimum value of b-a is 1

Solution: f(x)=1+x + f(x)=(1 x)+(x2 x3)+....x2012=（1﹣x）（1+x2+x4+…+x2010)+x2012 when x = 1, f (x) = 2 1006 + 1 = 2013 0, when x ≠ 1, f (x) = (1 x) (1 + x2 + x4+...+x2010）+x2012

1﹣x）•+x2012

0, f(x)=1+x + monotonically increasing on r;

f(0)=1, f(1)= 0, f(x)=1+x + has a unique zero point on (1,0), which is obtained by 1 x+3 0: 4 x 3, f(x+3) has a unique zero point on (4, 3) g(x)=1 x+ +g (x)=(1+x)+(x2+x3)+....x2012=﹣[（1﹣x）+（x2﹣x3）+…x2012]= f (x) 0, g(x) is monotonically decreasing on r;

and g(1)=( )0,g(2)=1+( n 2, = 0,g(2) 0

g(x) has a unique zero point on (1,2), which is obtained by 1 x 4 2: 5 x 6, and g(x 4) has a unique zero point on (5,6).

The function f(x)=f(x+3) g(x 4), the zero point of f(x) is the zero point of f(x+3) and g(x 4) The zero interval of f(x) is ( 4, 3) (5,6) and b,a z,(b a)min=6 ( 4)=10

Therefore, C