It is known that the quadratic function f x satisfies f 0 0 and f x 1 f x x 1, g x 2f x x.

Let f(x)=ax squared + bx+c,,, because f(0)=1, substituting 0 into c=0, that is, f(x)=ax squared + bx, and because f(x+1)=f(x)+x+1, substituting f(x)=ax squared + bx into this equation, we get ax squared + (b+1)x+1=ax squared + (2a+b)x+a+b, and the solution of a=b=b+1 from the principle of constant eqation gives a=b=one-half. Therefore, f(x) = one-half x square minus one-half x I believe that the landlord has done the second question, and I wish the landlord learning progress.

When x=0, f(1)=1 can be obtained, and when x=1, f(2)=3 can be obtained, from which we can find that the analytical formula of f(x) is f(x)=x 2 2+x 2, and f(-x)=x 2 2-x 2, so g(x)=x 2 It is not easy to look at the claw machine.

Let f(x)=ax +bx+c

f(0)=0, so c=0

f(x+1)=a(x+1)²+b(x+1)=ax²+2ax+a+bx+b=f(x)+x+1=ax²+bx+x+1

2ax+a+b=x+1

Therefore there is 2a = 1

a+b=1, so a=1 2, b=1 2

f(x)=x²/2+x/2

g(x)=2f(-x)+x=x²-x+x=x²f(x²)=x^4/2+x²/2

I wish the landlord learning and progress o(o

begging

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

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

The quadratic function is solved as f(x)=ax +bx+c

f(0)=0

c=0f(x+1)-f(x)=x+1 - this one, right?

a(x+1) +b(x+1)-ax -bx=x+1, i.e., ax +2ax+a+bx+b-ax -bx=x+12ax+a+b=x+1

2a=1，a+b=1

a=1/2,b=1/2

f(x)=1/2x²+1/2x

g(x)=2f(-x)+x=2[1/2x²-1/2x]+x=x²-x+x=x²

f(g(x))=1/2(x²)²1/2x²=1/2x^4+1/2x²

No, you said f(0)=0 followed by.

And f(x+1)=f(x)=x+1, then it is f(0)=1.

f(x)=ax²+bx+c

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

c=0f(x)=ax²+bx

f(x+1)=a(x+1)²+b(x+1)=ax²+(2a+b)x+(a+b)

f(x)+x+1=ax +(b+1)x+1, so ax +(2a+b)x+(a+b)=ax +(b+1)x+1, so the coefficient of x is the same as the constant term.

2a+b=b+1

a+b=1, so a=b=1 2

So f(x)=x 2+x 2

Solution: Let x=-1 and 0 be substituted into the equation to get f(-1)=0 f(1)=1, so x=-1 and x=0 are the two intersections of the quadratic function and the x-axis, so the function can be y=ax(x+1) substituting f(1)=1 into the equation to get a=1 2, so y=1 2x(x+1)=1 2x 2+1 2x

Answer dear --- to f'(x) integral, you can get f(x)=2 times the root number under (x+1)+c f(0)=1 substitution, get c=0, so f(x)=2 times the root number under (x+1).

The numerator of the derivative of this question is not x.

f(0)=1 is substituted to get c=0, so f(x)=2 times the root number (x+1).

The result of the question is incorrect.

Answer dear --- to f'(x) integral, you can get f(x)=2 times the root number under (x+1)+c f(0)=1 substitution, get c=0, so f(x)=2 times the root number under (x+1).

The result of the question is incorrect.

Dear ---1) Define the domain: Because the denominator x ≠0, the domain is defined as.

Range: f(x)=x+1 x-1, when x>0, using the inequality property x+1 x 2, if and only if x=1 x, i.e., x=1, the equal sign holds. At this point, f(x) 2-1=1

When x0) starts, the monotonicity definition proves that the function f(x) is a subtraction function over the interval (0, a under the root number) and an increasing function on the interval (a, + under the root number).

There is also the question of whether the expression is f(x)=x+1 (x-1)If this is the case, it should be written as f(x)=(x-1)+1 (x-1)-1, and use the commutation idea to make t=x-1, and return to the original way, that is, the definition domain has changed.

Let f(x)=ax2+bx+c, because f(0)=0, so c=so, b+1=2a+b, a+b=1, so a=1 2, b=1

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

f(0)=f(-1)-1+1

f(-1)=0

Let the double function be f(x)=ax 2+bx+c

c=00=a-b

1=a+ba= b= c=0

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

f(0)=f(-1)-1+1=0 f(-1)=0f(x) is a quadratic function f(0)=f(-1)=0, and the axis of symmetry is x=-1 2

Let f(x)=a(x+1 2) 2+b

f(1)=f(0)+0+1=1=9/4a+bf(0)=1/4a+b=0

The solution yields a=1 2

b=-1/8

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

f(x)=x^2/2+x/2

Solution: Since f(x)= is a quadratic function, f(x)=ax 2+bx+cf(0)=0, we know c=0

f(1)=0+0+1=1，a+b=1 ①

f(2)=1+1+1=3, 4a+2b=3 ②②2*①=2a=1, a=1/2

b=1-1/2=1/2