Let the quadratic function y f x satisfy f 0 1, and f x 1 f x 1 4x, find the analytic formula of f

Solution: Let the analytic formula of the quadratic function be f(x)=ax 2+bx+cf(0)=1

So c=1 f(x)=ax 2+bx+1

f(x+1)-f(x-1)=4x

f(1+1)-f(1-1)=4

f(2)=5

f(-1+1)-f(-1-1)=-4

f(-2)=5

4a+2b+1=5

4a-2b+1=5

b=0a=5/4

f(x)=x^2+1

Let f(x)=ax 2+bx+c, because f(0)=1, then c=1f(x+1)=a(x+1) 2+b(x+1)+1a(x 2+2x+1)+bx+b+1

ax^2+（2a+b）x+a+b+1

f(x-1)=a(x-1)^2+b(x-1)+1a(x^2-2x+1)+bx-b+1

ax^2+（b-2a）x+a-b+1

So f(x+1)-f(x-1)=4ax+2b, and because f(x+1)-f(x-1)=4x

So a=1, b=0

So f(x)=x 2+1

Let f(x)=ax 2+bx+c be brought in and it will be found.

Let f(x)=ax 2+bx+c be observed when f(0)=1, and the solution is c=1

From f(x+1)-f(x)=4x to obtain a(x+1) 2+b(x+1)-ax 2-bx=4x, (2a-4)x+(a+b)=0, for any x is destroyed, 2a-4=0, a+b=0

Solve the defeat macro sum to get a=2, b=-2

f(x)=2x^2-2+1.

Solution: Let the quadratic function f(x) ax 2 bx c,a≠0;

then f(x+1) a(x+1) 2 b(x+1) c;

From f(0)3, we get c3

f（x+1）-f（x）＝4x

A (Ax Judgment x +1) 2 B (Punch x + 1) C-(ax 2 bx C) 4x, 2ax a b 4x, then 2A 4, A 2, Silver not a b 0, b -2

Quadratic function f(x) 2x 2-2x+3

Let f(x)=ax 2+bx+c

Because f(0)=1, c=1

Because: f(x+1)-f(x)=2x+1, f(x+1)=f(x)+2x+1

When x=0, so f(1)=f(0)+1=1+1=2, when x=1, f(2)=f(1)+2+1=5, we get: a+b+1=2,4a+2b+1=5

So a=1, b=0

f(x)=x^2 +1

y=f(x) is a 2th order function.

Let f(x)=ax 2+bx+c

f(0)=1==>c=1

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

then there is a(x+1) 2+b(x+1)+1-ax 2-bx-1=2x2ax+a+b=2x

Equation identity. Then there is 2a=2 ,a+b=0 ==>a=1,b=-1, so f(x)=x 2-x+1

Note: Because in x different cases.

There is f(x+1)-f(x)=2x

After simplification, there is 2ax+a+b=2x

To make 2ax+a+b=2x true.

Only when 2a=2, a+b=0, 2ax+a+b=2x is true, x is any value, that is, the solution is 2a=2, a+b=0, and we get a=1, b=-1

Let f(x)=ax 2+bx+c

f(0)=0 is substituted into f(x)=ax 2+bx+c according to f(x+1)-f(x)=2x=2x, respectively.

3 equations are obtained: 4a+2b+c=3

0+0+c=1

a+b+c=1 just solve it and get a=1 b=-1 c=1.

f(x)=x 2-x+1 Try to keep it simple, not that complicated.

Bring x=-1 into f(x+1)=f(x)+x+1 Solve f(-1)=0 Bring x=0 into f(x+1)=f(x)+x+1 to solve f(1)=1 Then there are 3 points (0,0) (1,0) (1,1) Then let the analytic formula and bring in to solve the solution: a=1 2 b=1 2 c=0 f(x)=1 2x +1 2x

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What about the answer. It is known that f(x) is a quadratic function, if f(0)=0, and f=f+x+1Find the expression for f.

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Hello dear, the answer is y

Questions. My dear.

It was f(x) that was asked

The solution to this problem is:

Use the condition f=f+x+1 known to the problem.

You can calculate the value at x 1, x = 0, x -1, and then you can find it.

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Since f(x) is a quadratic function, let .

f(x)=ax^2+bx+c

f(0)=8

f(0)=c=8

f(x+1)-f(x）

a(x+1)^2+b(x+1)-(ax^2+bx)=2ax+(a+b)

2x+1 so has: 2a=-2

A+B=1 Combined (1)(2) launched.

a=-1，b=2

So f(x)=-x 2+2x+8

f(x)=-x^2+2x+8

(x-1)^2+9<=9

y=log3(f(x))

log3(9)=2

So the range of y is (negative infinity, 2].

Let f(x)=ax 2+bx+c(a≠0).

f(0)=c=0

f(x+1)=a(x+1) 2+b(x+1)=ax 2+bx+x+1 then, 2ax+a+b=x+1

The corresponding coefficients are equal: a = 1 2, b = 1 2

Then, f(x)=1 2 x 2 +1 2 x=1 2(x 2+x).

Solution: 1 set.

y=f(x)=ax²+bx+c

then f(x+1)-f(x-1)=...=4ax+2b=4xCompare the coefficients on both sides, there is.

4a=4, 2b=0

Therefore a=1 b=0

f(0)=c=1

Therefore f(x)=x +1

2 f(-x)=x²+1=f（x）

Therefore, it is an even function.

3 ymin=f(0)=1

ymax=f(2)=5

1.The quadratic function y=f(x) satisfies f(0)=1, let f(x)=ax 2+bx+1, f(x+1)-f(x-1)=4ax+2b=4x, and the comparison coefficients are 4a=4,2b=0,a=1,b=0.

f(x)=x^2+1.

2.Obviously f(-x) = f(x) and f(x) is an even function.

f(2)=5>f(-1), and the range of f(x) on [-1,2] is [1,5].

1 Let the quadratic function y=ax 2+bx+c c=0 because f(0)=0 so c=0

a(x+1)^2+b(x+1)-a(x-1)^2-b(x-1)=4xa(2x+2)+2b=4x

2ax+2(a+b)=4x

a=2 b=-2

So f(x)=2x2-2x

2.f(-x)=2x 2+2x≠f(x) f(-x)=2x 2+2x≠-f(x) So the function is non-odd and non-even.

When x=1 2, the minimum value of the function is -1 4

The maximum value f(-1)=f(2)=17 4

(1) Let f(x)=ax 2+bx+c, because f(0)=1, so c=1, because f(x+1)-f(x-1)=4ax+2b=4x, so a=1, b=0

f(x)=x^2+1

2) f(-x)=x 2+1=f(x), so f(x) is an even function (3) the range is [1,5], the minimum value = f(0)=1, and the maximum value = f(2)=5

From f(0)=1, when x=-1, there is f(0)-f(-2)=-4 to get f(-2)=5, and similarly when x=1, there is f(2)-f(0)=4

f(2)=5

Let the quadratic function y=f(x)=ax 2+bx+c, and f(0)=1

f（-2）=5

f(2)=5 substitution, get.

a=1, b=0, c=1, so there is.

f（x）=x^2+1；

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

Apparently an even function;

f(x)=x 2+1, we know that when x=0, there is a maximum value of 1, and f(2)=5, so the value range on [-1,2] is [1,5].

1.Let f(x)=ax 2+bx+c, because f(0)=1, so c=1, and because f(x+1)=a(x+1) 2+b(x+1)+c, f(x-1)=a(x-1) 2+b(x-1)+c, so f(x+1)-f(x-1)=4ax+2b=4x, so a=1, b=0, c=1

So f(x)=x 2+1

is an even function, symmetrical with respect to the y-axis, with the opening pointing upwards. So the maximum value of f(x) is f(2)=5, the minimum value is f(0)=1, and the range is [1,5].Complete.