Let the quadratic function f x x 2 bx c

Draw a sketch according to the topic and study the properties.

1 is the left zero point of the function.

The right zero point is greater than or equal to 3

So the axis of symmetry is greater than or equal to 2

Then look at the small question.

The first question is to directly substitute zero point 1.

In the second question, replace b with c according to the conclusion of the first question, and then write the axis of symmetry and look at the range.

The third question, sina is [-1,1], look at the monotonicity by yourself, and then directly write down the maximum value of algebra to solve b and c

Solution: (1) A 0, c 0 are obtained from the condition, and the discriminant formula =b 2-4ac silver mark-4ac 0 is obtained, so that the equation ax 2+bx+c=0 has two unequal real roots

2) Let g(x)=f(x)-[1 2][f(x 1 )+f(x 2)], prove g(x 1) g(x 2) 0, and g(x)=0 must have a solid root in (x 1, x 2), and the problem is proved

Proof of: (1) f(1)=0, a+b+c=0 and a b c, a 0, c 0, i.e., ac 0

again =b2-4ac -4ac 0, the equation ax2+bx+c=0 has two unequal solid roots in the cavity

Therefore, the function f(x) must have two zeros

2) Let g(x)=f(x)-[1 2][f(x1)+f(ante and x2)], then g(x1)=f(x1)-[1 2][f(x1)+f(x2)]=

f(x1)−f(x2)

2，g（x2）=f（x2）-[1/2][f（x1）+f（x2）]=

f(x1)−f(x2)

2，g（x1）•g（x2）=

f(x1)−f(x2)

f(x2)−f(x1)

2=-[1/4][f（x1）-f（x2）]2．

f（x1）≠f（x2），∴g（x1）•g（x2）＜0．

g(x)=0 must have a solid root in (x1, x2).

The equation f(x) = [1 2] [f(x1) + f(x2)] must have a real root in (x1, x2).

Then from g(x1) g(x2) 0, we can get that the function value of the quadratic function g(x) can be positive or negative, so the function g(x)=f(x)-[1 2][f(x1)+f(x2)] must have two intersections with the x-axis, so the equation f(x)=[1 2][f(x1)+f(x2)] has two unequal real roots

In summary, the equation f(x)=[1 2][f(x1)+f(x2)] has two unequal real roots, and there must be one real root belonging to (x1, x2).

6,1) a+b+c=0 a>b>c Therefore a>0 c<0 b is uncertain but dert>0

2) Concave function properties are plotted.

f(x)-1 2[f(x1)+f(x2)]=0 There are two unequal real roots, so dert>0 Let x3 belong to (x1,x2),2, and the known quadratic function f(x)=ax 2+bx+c

1) If a b c and f(1)=0, it is proved that f(x) must have two zeros;

2) If there are two unequal real roots for x 1, x 2 r and x 1 x 2, f(x 1) ≠ f(x 2), the equation f(x) = [1 2] [f(x 1) + f (x 2)] proves that there must be one real root belonging to (x 1, x 2).

1) Proof of two known inequalities.

In the average order x=2, get.

2≤f(2)≤4²/8=2

2) Solution: f(2)=4a+2b+c=2, f(-2)=4a-2b+c=0 solution concealment obtains b=1 2, c=1-4a

f(x)-x=ax +(b-1)x+c=ax -(1 2)x+1-4a 0 is constant, i.e.

a 0, =1 2) -4a(1-4a)=(4a-1 2) 0 solution punches with friends a=1 8, where f(x)=(1 8)x +(1 2)x+(1 2)=(1 8)(x+2) satisfies another known inequality (" is " " or " = "scattered).

Therefore f(x) = (1 8) x + (1 2) x + (1 2).

Solution: Answer: (-infinity, -2】u[4,+infinity).

When x belongs to [1,2], f(x)|>=x <==> |f(x)/x| >=1 <==>|x+c/x+b|>=1

The constructor g(x)=x+c x+b, x belongs to [1,2], and the maximum value of the function g(x) is m, and the minimum value is m, then g(x)|Maximum = max

The original problem <==> for any real number b, m|>=1 or m|>=1, find the range d for c. <==> inequality m|with respect to bThe union of >=1 and m>=1 is r, and the range of the real number c is obtained.

If c<=0 then g(x) is an increasing function, so m=1+c+b , m=2+c 2+b, and if 01 then g(x) in the interval [1, root number c] is a subtraction function, and in the interval [root number c,2] is an increasing function.

m=2 root number c+b, m=max, 12, m=1+c+b;

i) If c<=1, m|>=1 ===>|1+c+b|>=1 <==> b>=-c or b<=-2-c

m|>=1 <==>|2+c/2+b|>=1 <==> b>=-1-c2 or b<=-c2-3

Remember the set a=, b=

aub=r, using the number axis, the number form is combined, c<=1===>-1-c 2<-c, -3-c 2<-2-c

When -1-c 2<=-2-c i.e. c<=-2, aub=r

ii), when c>1, the maximum value of g(x) = max}

Let c=, just satisfy aubuc=r.

There is a real number x0 [1,2] such that |f(x)|x Establishment Instructions are only required|f(x)|The maximum value is greater than or equal to x

The maximum value is only taken at the endpoint.

Then only lf(1)l>=1,|f(2)|>=2 so |1+b+c|>=1,|4+2b+c|>=2

Let's teach the method, **b on r (segmented value) to find the value of c.

Since f(x) is an even function, the property of the even function is that f(x)=f(-x), and f(-x)=(-x) 2-bx+c; So -bx=bx, so b=0; As for 1+c=0, it is the condition f(1)=0, and it is 1 2+c=0, so c=-1; So, f(x)=x 2-1

Prove that x0 (x1,x2) exists so that f(x0) = 1 2[f+f] holds.

It is enough to prove that [f(x1)+f(x2)] 2 is a function value of f(x) over the interval (x1, x2).

f（x）=ax^2+bx+c．

Let a>0 and let its axis of symmetry be x=h and the vertex coordinates be (h,k) when hx2, the same can be shown that when x1 h x2, if f(x1)>f(x2), the range of f(x) is [k,f(x1)].

Then [f(x2),f(x1)] is the sub-Luchang interval of the value range.

f(x1)+f(x2)] 2 [f(x2),f(x1)]f(x1)+f(x2)] 2 [k,f(x1)] if f(x1)+f is accompanied].

The first problem, a system of simultaneous equations, solves x=-c a under the root number, or =-c aa1b1=2 -c a under the root number

The equation is listed from a>b>c, a+b+c=0, and the equation is obtained under the root number 1 2 under the root number -c a root number 2

So the root number 2 a1b1 2 times the root number 2

The second a is a natural number, 0 is not good, so 1 is the smallest, because the shape of the function of a = 1 has been determined, move the function image to tangent to the x-axis, the tangent point is between 0 and 1, and then move down a little more to appear two unequal positive roots less than 1.