Knowing that the quadratic function f x satisfies f 0 1 and f x 1 f x 2x, find f x

f(0+1)-f(0)=0, f(1)=f(0)=1, f(1+1)-f(1)=2, f(2)=3, you can get (0,1),(1,1),(2,3) three points set f(x)=ax 2+bx+c, and it is easy to get the answer f(x)=x 2-x+1Topics like this can make full use of the known conditions, although the questions give few conditions, in fact, it is basically a substitute thing, or you can draw a picture to help you solve the problem more vividly. In mathematics, you need to learn how to combine numbers and shapes, and some problems are basically a matter of drawing a diagram.

Let the quadratic function f(x)=ax 2+bx+c, then f(x+1)=a(x+1) 2+b(x+1)+c

f(x+1)-f(x)=a(2x+1)+b=2ax+(a+b) 2xa=1, a+b=0 i.e. b=-1

c=1 is obtained from f(0)=1

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

Let f(x)=ax square + bx+c, let x=0, solve c=1, and then use the formula given in the problem to make x=1, and solve f(1)=1. Knowing that f(x)=ax squared + bx+1, bring in x=1 to solve a+b=0, b=-a, then f(x)=ax squared - ax + 1, using the formula a(x+1) square - a(x+1)+1-ax square + ax-1=2x, the solution is a=1, so the equation is x's square - x+1

f(x)=ax^2+bx+1

f(x+1)=a(x 2+2x+1)+b(x+1)+1. 2ax+a+b

Set by the question. f(1)=1,f(2)=3

So a+b=0

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

We can let f(x)=ax+b pass the first condition b=1, and the second condition into the equation a=2x, so the equation is f(x)=2x squared + 1

Solution: (1) Let f(x)=ax +bx+c

f(x+1)-f(x)

a(x+1)²+b(x+1)+c-ax²-bx-c=2ax+a+b

That is, 2ax + a + b = 2x

So 2a=2, b+a=0, i.e. a=1, b=-1f(0)=c=1

Therefore, the image of f(x)=x -x+1=(x-1 2) +3 4(2) in the range [3 4,3] y=f(x) on the interval [-1,1] is always above y=2x+m.

then x -x+1>2x+m i.e. x -3x+1-m>0 Hengcheng Idiot = 9-4(1-m)<0

Solution: m<-5 4(2).

(1) From f(0)=1, there is f(1)-f(0)=0==> f(1)=f(0)=1

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

From f(0)=1 there is c=1

From f(1)=1 there is a+b+1=1==>a+b=0f(x)=ax 2-ax+1

f(x+1)=a(x+1) 2-a(x+1)+1f(x+1)-f(x)=a(2x+1)-a=2x==>a=1, then f(x)=x 2-x+1

2) To make the line below f(x), then for the version-1 x 1 satisfies x 2-x+1>2x+m

m at y=(x-3 2) 2-5 4 decreasing.

When x=1, the minimum weight is 1 4-5 4=-1

then m<-1

(1) Let the complex f(x)=ax system 2+bx+c, then f(x+1)-f(x)=2ax+a+b=2x

a=1;b=-1

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

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

2) Since the f(x) image is always above the line y=2x+m, then f(x)=y=2x+m => x 2-3x+(1-m)=0 discriminant δ

0 => m<-5/4

Let f(x)=ax +bx+c

f(0)=1.

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

i.e. a[(x+1) -x ]+b[(x+1)-x]=2x2ax+a+b=2x

So a=1b=-1

So f(x)=x -x+1

Since 2) take into account the function image opening up bai

Simultaneous y=x -x+1

y=2x+m

We get x -3x+1=m, x du[-1,1], i.e., -1 zhim 5 when the two functions have an intersection point.

DAO so m<-1 or m>5

(1) Order.

baif(x)=ax2+bx+c(a≠0) is substituted into duf(zhix+1)-f(x)=2x, dao obtains: version a(x+1)2+b(x+1)+c-(ax2+bx+c)=2x,2ax+a+b=2x,f(x)=x2-x+1;

2) When x [-1,1], f(x) 2x+m is established, i.e., x2-3x+1 m is established;

Let , x [-1,1] then the axis of symmetry: , then g(x)min=g(1)=-1

m≤-1；

f(0)=1, let f(x)=ax 2+bx+1f(x+1)-f(x)=a(2x+1)+b=2ax+a+b=2x contrast coefficients: 2a=2, a+b=0

i.e. a=1, b=-1

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

1）f(x)=f(x)-g(x)=x^2-(m+1)x-1=[x-(m+1)/2]^2-1-(m+1)^2/4

The axis of symmetry is x=(m+1) 2

If the axis of symmetry is in the interval, i.e., -3=3, f(m)=f(2)=1-2m, if the axis of symmetry is on the left side of the interval, i.e., m<-3, f(m)=f(-1)=m+12)m [-1,2], f(m)=-1-(m+1) 2 4, its minimum value is when m=2, fmin=-13 4

Let f(x)=ax +bx+c

f(x+1)-f(x)

a(x+1)²+b(x+1)+c-ax²-bx-c=2ax+a+b

That is, 2ax + a + b = 2x

So 2a=2, b+a=0, i.e. a=1, b=-1f(0)=c=1

So f(x)=x -x+1=(x-1 2) +3 4 is in the range [3, 4, 3] on the interval [-1,1].

The image of y=f(x) is always above y=2x+m.

Then x -x+1>2x+m, i.e., x -3x+1-m>0 is constant, =9-4(1-m)<0

Solution m<-5 4

f（x）=ax^2+bx+c

f(0)=0+c=c=1

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

f(x+1)-f(x)=2ax+(a+b)=2x2a=2,a=1

a+b=0,b=-a=-1

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

2x+mm let g(x)=(x-3 2) 2-5 4 be a monotonic reduction function in [-1,1], so the minimum value is g(1)=-1

So, the range of m is m<-1

Let f(x)=ax +bx+c

Because f(0)=1, c=1

So f(x)=ax +bx+1

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

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

So 2a = 2 and a + b = 0

So a=1b=-1

So f(x)=x -x+1

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

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

f(0+1)-f(0)=0 i.e.: f(1)=f(0)=1f(1+1)-f(1)=2 i.e.: f(2)=f(1)+2=3 Let this quadratic function be f(x)=ax 2+bx+c then there is:

c=1a+b+c=1

4a+2b+c=3 Solution: a=1, b=-1, c=1, so there is:

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

f(x^2)=x^4-x^2+1

Solution: Let f(x)=ax +bx+c

Substituting f(0)=1 gives c=1

So f(x)=ax +bx+1

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

Substituting the above two formulas into f(x+1)-f(x)=-2x-1 gives ax +(2a+b)x+a+b+1-ax -bx-1=-2x-12ax+a+b=-2x-1

The contrast coefficient yields 2a=-2

a+b=-1

The solution yields a=-1 b=0

So f(x)=-x +1

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

Because f(0)=1, c=1

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

So a(x+1)2+b(x+1)+1-(ax2+bx+1)=-2x-1

2ax+a+b=-2x-1

So a+b=-1

2a=-2, so a=-1

b=-1-a=-2

So f(x)=-x2-2x+1

Oh ie body red don't send oh ie five years later, so late still desperately running outside.