The minimum value of the function y x a 2 x b 2 a b is a constant is

y＝(x－a) ^2＋(x－b)^2

x²-2ax+a²+x²-2bx+b²

2x²-2（a+b）x+a²+b²

There is a maximal formula for a quadratic function that yields:

its minimum value. 【4×2（a²+b²）-4（a+b）²】/8（a²+b²-2ab）/2

a-b）²/2

1. If a=b, then the minimum value of y = 0, 2, if a is not = b, then y=(x a) 2 (x b) 2=2x 2-2(a+b)x+(a 2+b 2), the discriminant formula of the root <0, the parabola with the opening facing up, there is no intersection with the x-axis, the axis of symmetry x=(a+b) 2, then the lower vertex of the parabola is the minimum value y=(a-b) 2 2

Suddenly, the comrades in front of me were really fast.

The form of the quadratic function y=2x 2-(2a+2b)x+a2+b 2

Since the coefficient of the quadratic term is 2>0, the x value of the axis of symmetry of the function reflects the minimum value of the function. The axis of symmetry is x=(a+b) 2, just substitute it in, the answer will not be written, very slippery sorry Chang coarse, I wish you good luck in Nai Rang Town.

y Chong Jianzheng (x a).

2＋(x－b)^2

x²-2ax+a²+x²-2bx+b²

2x²-2（a+b）x+a²+b²

Yes. Quadratic functions.

The most valuable formula for dissipation can be repented to the wanton:

its minimum value. 【4×2（a²+b²）-4（a+b）²】8（a²+b²-2ab）/2

a-b）²/2

construct vectors m (1,1),n (a-x,x-b).

Vector modulus inequality |m|·|n|≥|m·n|, get (1 +1 )[a-x) +x-b) ] 1· (a-x)+1· (x-a)]

a-x)²+x-b)²≥a-b)²/2.

Therefore, the minimum value is obtained: (a-b) 2

At this point, a-x x-b, i.e., x(a+b)2.

Parse: f( x )=x - a |+x - b |Band grip|( x - a )-x - b )|a - b |.

Answer: Stupid Changqing Xun Zao: | a - b |.

Solution: 1) Proof is as follows:

Derivative of f(x)=(ax+b) (x 2+1), f'(x)=[a*(x^2+1)-(ax+b)*2x]/(x^2+1)

a(x^2+2b/a*x-1)/(x^2+1)^2， ①

Because a>0,f'The symbol of (x) is determined only by the symbol of x 2+2b a*x-1 (as opposed to).

Order f'(x)=0 yields:-a(x 2+2b a*x-1) (x 2+1) 2=0, i.e., a(x 2+2b a*x-1)=0,x 2+2b a*x-1=0,

The discriminant formula of this quadratic equation δ=4(b a) 2+4>0 (constant is positive), so there must be two real roots. That is, the quadratic function x 2+2b a*x-1 must have two zeros, and near the two zeros, the quadratic function changes signs, and it is easy to know from the image of the quadratic function, and the signs near the two zeros change in opposite directions (one zero point accessory symbol changes from positive to negative, and the other zero near the symbol symbol changes from negative to positive). Hence f'(x) has only two zeros, and the signs near the two zeros change in the opposite direction, then f(x) takes the maximum value and the other takes the minimum value at the two stationary points.

2) Let the solution of the equation be: x1, x2, and x10, so b 2+a 2≠0, then 4-a 2=0, a=2(a>0).

Substituting a=2 yields: b[16+8-2*b 2]=0,

So b=0 or 2 3.

Since the equation is not a sufficient and necessary condition for the equation, it needs to be substituted for verification.

When a=2, b=0, by , x1=-1, x2=1, f(x1)=-1, f(x2)=1, meet the requirements;

When a=2,b=2 3,,x1=( 3+1) 2 2,x2=-( 3-1) 2 2,assuming x1 when a=2,b=-2 3, ,x1=( 3-1) 2 2,x2=-( 3+1) 2 2, and hypothetical x1 a=2,b=0 is the result.

Seek the derivative first, so that the derivative function is 0, and if the equation has different real roots, then question 1 is true.

Knowing that 1 and -1 are the roots of the equation from 2, we can write the equation in a question as (x-1)(x+1)=0, and we can get a=1

1. Segmentation of segmentation functions.

Second, the application of basic inequality.

For reference, please smile.

Solution: According to Cauchy's inequality:

y=（x-a）^2+(b-x)^2

(x-a+b-x) 2] 2 = (b-a) 2 If and only if x-a = b-x, the equal sign holds, then: x = (a+b) 2So.

y ≥ b-a)^2

The minimum value is: (b-a) 2

The form y=2x 2-(2a+2b)x+a 2+b 2 , because the coefficient of the quadratic term is 2>0, so the x value of the symmetry axis of the function reflects the minimum value of the function. The axis of symmetry is x=(a+b) 2, just substitute it, and the answer will not be written, sorry, good luck.

The function becomes y=2x -2(a+b)x+a +b The parabolic opening is upward, and the minimum value is vertex coordinates 8=(a-b) 2

y =2x 2-(2a+2b)x +a2+b 2, then the point of the minimum value can be known by using the quadratic function is x =(a +b) 2, so the minimum value is [(a -b) 2] 2

The minimum value is 0 because the square of any number is greater than or equal to 0, so it is 0

(x－a)^2 ＋(x－b)^2 ≥(x-a-x+b)^2/2=(b-a)^2/2

Important inequalities.

a+b)/2≤√（a^2+b^2)/2)a^2+b^2≥(a+b)^2/2

In the above equation, x-a is considered as a, and b-x is regarded as b).