The function f x 1 is an even function, and when X 1, f x x 2 1, when X 1 is found, the expression o

The correct answer should be f(x)=x 2-4x+5

f(x+1) is an even function, so f(-x+1)=f(x+1); This shows a new conclusion: the f(x) image is symmetrical with respect to the straight line x=1, and when x>1, -x<-1==>-x+2<1 f(-x+2)=(-x+2) 2+1=x 2-4x+5 f(-x+2)=f[-(x-1)+1]=f[(x-1)+1]=f(x) i.e.: f(x)=x 2-4x+5 (x>1) Description:

There are three difficulties in this question: 1Even function is converted to symmetry 2 Why add 2 3 after multiplying the variable by minus one, the application of f(-x+2).

f(x)=x 2-4x+5 Analysis: Because f(x+1) is an even function. So f(x+1) = f(-x+1), so f(x) = f(2-x).

Let x>1, then 2-x<1 so f(x)=f(2-x)=(2-x) 2+1. In summary: when x>1, f(x)=x 2-4x+5

Dizzy. Is it supposed to be"Because f(x+1)=x2+1 at x<1, f(x-1+1)=(x-1) 2+1i.e. f(x)=x2-2x+2

And because f(x+1) is an even function, it can be seen as f(x) translating one unit to the left, so the parity does not change, so when x>1, then -x<1, f(-x)=(-x) 2-2x+2=f(x)So f(x)=x 2-2x+2"..

Kindness. The above is purely a personal opinion, but I will check it out.

If it's not right, I'll come back and correct the mistake.

It is possible to make f(x) = f(x+1), which is actually saying that f(x) is an even function, so f(-x) = f(x), i.e. f(x+1) = f(-x+1).

f(x+1) is an even function, not that f(x) is an even function, so you have to construct a function with (x+1) as a parameter. The condition of the even function can be applied.

The expression for f(x) is f(x)=x 2+1, and when finding x r, the even function is symmetric with respect to the y-axis, so f(x)=f(-x) just change x.

Because f(x+1) is an even function, f(-x+1) = f(x+1), i.e., f(x) = f(2-x);

When x 1, 2-x 1, at this time, f(2-x) = (2-x)21, i.e. f(x) = x2

4x+5, so the answer is: x2

4x+5．

f(x+1) is an even function.

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

The axis of symmetry is x=1

As long as you find the vertex of f(x)=x2+x, the bending and the symmetry point of x=1 can be obtained analytically.

When x 1 f(x)=x 2+x=(x + 1 2) 2 - 1 4 vertices are (-1 2 ,-1 4), and the symmetrical points about x=1 are buried (5 2 ,-1 4).

So when x 1, f(x) = (x - 5 2) 2 - 1 4 = x 2-5x+6

f(x) is an even function xf(x) is an odd function so (1,1)xf(x)dx= 0

1,1)x[x+f(x)]dx= ∫1,1)x^2dx=x^3/3|(-1,1)=2/3

A brief analysis, the song is detailed and forgiving, as shown in the picture of the wild judgment.

f(x)=x(1+x)=x²+x

When x 0, f(x)=x2+x, so let x call 0, -x 0, then f(x)=f(lingkong-x)=(x) +x)=-x -x=x(x-1).

That is, when x<0, the expression f(x) is f(x) ulnar chain blind = x(x-1).

f(x+1) is an even function.

f(x) is symmetrical with respect to x=1.

For any x, there is a grip on the object, and when the pure f(x)=f(2-x)x>1, 2-x<1, f(x)=f(2-x)=(2-x) +1f(x) The expression is: when x<1, f(x)=x +1; At x>1, f(x)=(x-2) +1

Solution: Since the function f(x+1) is an even function, that is, f((-x)+1)=f(x+1), if x=1, then f(0)=f(2), we can know that the axis of symmetry of the function is x=1, then the image is also symmetrical with respect to the axis of symmetry, so when x<1, f(x)=x 2+1, when x>1, the expression f(x) is.

f(x)=（x-1）^2+1。

Take 2-x<1 and bring in f(x) to get f(2-x), because it is an even function, so f(x-2)=f(2-x).

Then take x'+2 instead of x, get the result.

x>1,f(x)= f((x-1)+1)

f(-(x-1)+1) - because the function f(x+1) is an even function.

f(2-x)

2-x) 2 + 1 -- because 2-x < 1

x^2-4x+5

Answer: f(x)=x 2 4x 5

Since: f(x 1)=(x 1) 2 1;x<0 and then known by the properties of the even function:

f(x＋1)=f(－x+1)=(－x＋1)^2＋1;When x>0 is replaced with y=x 1Get: f(y)=y2 4y5;y>1

f(x+1) is an even function, then f(x+1)=f(-x+1).

So the axis of symmetry is x=1

So when x>1, f(x)=(x-2) 2+1=x 2-4x+5