Ask for questions about the first year of high school mathematics, the notation of functions

Let me answer.

f(1)=-5

Because f(x+2)=1 f(x).

So f(1+2)=1 f(1) (

i.e. f(3)=-1 5

Similarly f(3+2)=1 f(3) (

i.e. f(5)=1 (-1 5)=-5

f(f(5))=f(-5) (

Now let's find f(-5).

f(-1+2)=1/f(-1) (

i.e. f(1)=1 f(-1)=-5

So f(-1)=-1 5

f(-3+2)=1/f(-3) (

So f(-3)=-5

f(-5+2)=1/f(-5) (

f(-5)=-1/5

by (@), so.

f(f(5))=-1/5

All of the above equations marked with (*) are because f(x+2)=1 f(x).

solution, because f(x+2)=1 f(x), so f(x+4)=1 f(x+2).

So f(x+4)=f(x), i.e. f(x) is a periodic function with period 4.

So f(5) = f(1) = f(-3) = -5

And because f(x+2)=1 f(x).

So f(-5)=1 f(-3)=-1 5

m=,n=then only one mapping from m to n should be a-1 or a-2 ("a-1" means that a corresponds to 1) There are two kinds, if m=, n= then only one mapping can be established from m to n, mainly because any element in the m to n map m set must correspond to one element in n, and the element in n can correspond to multiple elements in m, so there is only 1,2-a at this time, which is mainly determined by the nature of the function, For any value of the independent variable x, the strain y has a uniquely determined value to which it is opposed.

If you don't figure it out, you can think about the function, y=f(x), you must have only one y value when you use an x, but a y value may correspond to multiple x values, and the function is essentially a set of maps.

It's like polygamy in ancient times. The husband is n. The wife is m. A husband may have more than one wife, and a wife may have only one husband. Except for remarriage!!

1. The distance between the nuclear power plant and cities A and B shall not be less than 10km, so x 10 and 100-x 10

Define the domain as 10 x 90

The total monthly power supply fee y= (x * 20) + 100-x) * 10 = 2, y =

When a nuclear power plant is 100 km away from the A3 km, the electricity bill is minimal.

1.Let f(x)=ax +bx+c

From f(0)=1, c=1

f(x+2)-f(x)=4x+6

a(x+2) +b(x+2)+c-ax -bx-c=4x+6 4ax+4a+2b=4x+6 4a=4,4a+2b=6, i.e. a=1, b=1

Analytic formula f(x) = x +x+1

3.Functions are quadratic.

m-1≠0,m²+m=2

m + m-2 = 0, (m-1) (m+2) = 0 m = -2 or m = 1 (round).

i.e. m=-24∵f(x)=2f(1/x)+x

Replacing x with 1 x is: f(1 x) = 2f(x) + 1 x f(x) = 4f(x) + 2 x + x

3f(x)=2/x+x

f(x)=-1/3(2/x+x)

5.Let f(x)=ax+b

then f(f(x))=a(ax+b)+b=a x+(ab+b)f(f(f(x))))=a[a x+(ax+b)]+b=a x+a b+ab+b=8x+7

a³=8∴a=2

4b+2b+b=7,∴b=1

f(x)=2x+1

Let f(x)=ax2+bx+c, since f(x) conceals f(x+2)=f(x-2), so f(x) is symmetrical with respect to x2. Let the two bits a, b then a2 + b2 = 10

If there is f(x), the elephant (0,3) gives c=3

1.Knowing f( x-1)=x+1, the expression for f(x) is x-1=t x=1+t x=(1+t)2f(t)=(1+t)2+1

i.e. f(x) = (1+x)2+1

2.If you know the function f(x)=x2-4x+3, find f(x+1).

f(x)=(x-1)(x-3)

f(x+1)=(x+1-1)(x+1-3)=x(x-2)3.Knowing that the quadratic function f(x) satisfies f(0)=2, f(x+1)-f(x)=x-1, find the explanatory formula of f(x).

When x=0f(1)-f(0)=-1 f(1)=f(0)-1=1, when x=1

f(2)-f(1)=0 f(2)=f(1)=1 when x=2

f(3)-f(2)=1 f(3)=f(2)+1=2...Draw the pattern:

an=a(n-1)+(n-1)

The above equations are added together.

a1+a2+a3+--a(n-1)+an=a1+[a1+a2+a3+--a(n-1)]+1+2+3+--n-1)

an=1+[1+2+3+--n-1)]=1+(1+n-1)(n-1)/2

n²-n+2)/2

The nth is (n -n+2) 2

In the same way, we get x<0 (n -3n+4) 2, i.e., x<0 f(x)=(x -3x+4) 2x=0 f(x)=2

x>0 f(x)=(x -x+2) 2 I don't know what grade you go to Actually, there is an easy way to do it in college, and this is a stupid way.

I won't explain the first two, it's relatively simple to change the yuan.

3.Let the quadratic inner function f(x)=ax2+bx+cf(0)=2, so allow c=2, quadratic function f(x)=ax2+bx+2f(x+1)-f(x)=x-1

a(x+1)2+b(x+1)+2-[ax2+bx+2])=x-12ax+a+bx+b-bx=x-1

So 2a = 1

a+b=-1

a=b=so f(x)=

Solution: For the problems you listed, you might as well set (x+1) x=t, then x=1 (t-1) (

So (x 2+1) x 2=1+[1 x 2]=1+(t-1) 21 x=t-1

Substitute this ( ) formula into the original form.

f(t)=1+(t-1) 2+(t-1)=t 2-t+1, so f(x)=x 2-x+1

If you don't understand, please feel free to ask.

Because the necessary condition for a fraction to be meaningful is that the denominator is not equal to 0, i.e., x≠0, (x+1) x≠1. The final result should be f(x)=x 2-x+1(x≠0).

This depends on the range of (x+1) x because (x+1) x equals 1 is , x has no solution, so.

The definition field of f(x) does not contain 1

For example, if x+1 is 2 and x is also 2, the result will be equal to 1, and the condition of not equal to 1 will not be satisfied.

The place you circled itself is wrong, (x+1) x≠1, then the solution is x+1≠x, which is obviously wrong.

The answer is you solving it or is it swollen?

Let u=1-x 1+x

Denominator: u(1+x)=1-x

Shift: ux+x=1-u

Shift the term, mention the common factor: (1+u)x=1-u

So we get x=(1-u) (1+u), and substitute it into the original formula to get :

f(u)=(1-u)/(1+u)

Replace u with x to get f(x) = (1-x) (1+x).