It is known that the function f x satisfies f x f 1 e x 1 f 0 x 1 2 x

f(0)=f'（1）/e………1)

Derivative of f(x):

f'(x)=f'(1) e (x-1)-f(0) + x then f'(1)=f'(1)-f(0)+1

Then f(0)=1

then there is f according to (1).'(1)=e

So f(x) = e x-x + (1 2) x 2f'(x)=e^x+x-1

x>0,f'(x)>0;x<0,f'(x) <0 then f(x) decreases monotonically at (- 0); At [0,+ monotonically incrementing.

2）f（x）≥（1 /2）x²+ax+b

That is, g(x)=f(x)-[1 2)x +ax+b]e x-(a+1)x-b

If the above equation holds for - then :

a+1)≥0………2）

If (2) is true, then we know that g(x) is an increasing function on r.

g(0)=1-b>0

Then b>1

And according to (2) A+1 0

Then (a+ 1)b 0, i.e. the maximum value is 0

Note: The second question I did according to f(x) (1 2)x +ax+b constant establishment, your question is not clear

<>This is the answer given by the teacher

Summary. Since we know the function f[ (x)]=x +1=e to the power (x), we get the function (x)=ln(x+1).

If the function f(x)=e*,f[ (x)]=x +1, then (x)=known function f(x)=e*, f[ (x)]=x +1, then (x)=known function f(x)=e*, f[ (x)]=x +1, then (x)=e is an x but can't be typed, like x, is that still the answer?

Knowing the function f(x)=e*,f[ (x)]=x +1, then (x)=then you send me the little x on the top of the e, put it together like that, my phone can't call, you call me, I'll copy.

If the function f(x)=e*,f[ (x)]=x +1, then (x)=known function f(x)=e*, f[ (x)]=x +1, then (x)=known function f(x)=e*, f[ (x)]=x +1, then (x)=known function f(x)=e*, f[ (x)]=x +1, then (x)=known function f(x)=e*, f[ (x)]=x +1, then (x)=known function f(x)=e*, f[ (x)]=x +1, then (x)=

Summary. Hello, knowing the function f(x)=x -7, then f(-4)=,f(0)=The specific calculation process is as follows:

If the function f(x)=x -7 is known, then f(-4)=,f(0)=Hello, if the function f(x)=x -7 is known, then f(-4)=,f(0)=The specific calculation process is as follows:

f（-4）＝（4）²-7＝16-7=9

f（0）＝0²-7＝0-7=-7

You just need to substitute the corresponding x into it.

f(f(2))？

Because f(2) 2 -7 4-7=-3, so f(-3) (3) -7 2, so f(f(2)) 2

If (x-21)=3, then x?

x-21＝3x＝3+21=24

The known function f(x)=x+1 when x<0; f(x)=-x-1 when x 0; then the solution set of inequalities x(x+1)f(x-1) 3.

Solution: When x<0 f(x-1)=(x-1)+1=x, then x(x+1)f(x-1)=x(x+1) 3....1)

When x<-1, x+1<0, so x(x+1)<0<3, i.e., (1) is true; When x=-1, x+1=0, then x(x+1)=0<3 is true;

When -10, -x (x+1)<0<3, i.e., (2) is true; So for any x 0, no.

Equation (2) is constant.

The solution of the original inequality is the whole real number.

f(x)2f(1 x)x.

1 Shake brother hidden x instead. x

Get: f(1 x).

2f(x)1/x

Multiply 2 by both sides at the same time to get:

2f(1 dust residual x).

4f(x)2/x

And the original phase section hall minus:

3f(x)2xx so.

f(x)2/(3x)x/3

f(x+y)=f(x)f(y)

Let y=1f(x+1)=f(x)*f(1)=f(x) 2, which is the series of equal stool rolling ratios.

f(n)=(1/2)^n

an=(1/2)^n

The first n items and. a1+a2+..an=1-(1 2) n because (1 2) n>0

So a1+a2+.an<1

bn=nf(n+1) f(n)=n 2 is a series of equal differences.

sn=n(n+1)/4

1/sn=4/n(n+1)=4[1/n-1/n+1]1/s1+1/s2+..1/sn=4[1-1/2+1/2-1/3+……1/n-1/n+1]

4[1-1/n+1]

4n/(n+1)

However, I laughed and thought that the last question might be a bit problematic, and the rest of the pants was estimated to be bn=nf(n) f(n+1), so there would not be 4 times.

f(x+2)=f[(x+1)+1]=1/f(x+1)=1/[1/f(x)]=f(x)

So f(x) is a periodic function with period 2;

And because f(x) is an even function, when x [-1,0], f(x)=f(-x)=-x

In this way, the image of the function f(x) and the hail can be completely determined;

g(x)=f(x)-kx-k has four zeros, which is the equation: f(x)=k(x+1) has four roots;

Draw the image of the function f(x) and the straight line y=k(x+1), and investigate the slope change of the straight line embedded

On the interval [1,3], when k=0, there is an intersection point;

01 4, so the function g(x) cannot have four zeros;

Are you looking at the topic of demolition of the shed?