Given that a is a real number, the function f (x) = 2aX & # 178; + 2x-3 is the minimum value on the interval [- 1,1], which is g (a), find g (a) The analytic expression of G (a) is obtained

Given that a is a real number, the function f (x) = 2aX & # 178; + 2x-3 is the minimum value on the interval [- 1,1], which is g (a), find g (a) The analytic expression of G (a) is obtained

solution
When a = 0, f (x) = 2x-3, the minimum value on [- 1,1] is f (- 1) = - 5, that is, G (0) = - 5;
When a > 0, the opening of function f (x) is upward, the axis of symmetry is a straight line, x = - 1 / (2a) is on the left side of Y-axis and intersects at (0, - 3) point of y-axis
If - 1
a=0 g=-5
When a is not equal to 0
f(-1/2a )=1/2a-1/a-3=-1/2a-3
f(-1)=2a-5
f(1)=2a-1>2a-5
a> 0 hour
The axis of symmetry has a > 1 / 2 in the interval
g=-1/2a-3
A
There is such a problem: when x equals one fourth and Y equals 2013, find the polynomial 7x ^ 3-6x ^ 3Y + 3x ^ 2Y + 3x^
XiaoCong said that the condition x is equal to one fourth and Y is equal to 2013 is redundant. Is his statement reasonable? Why?
Is it 7x ^ 3-6x ^ 3Y + 3x ^ 2Y + 3x ^ 3 + 6x ^ 3y-3x ^ 2y-10x ^ 3 + 3?
The combined result is 3, which has nothing to do with the value of X and y
His explanation is right. It really makes sense
Is there such a nonnegative integer m that the quadratic equation m2x2 - (2m-1) x + 1 = 0 with respect to X has two real roots? If it exists, request the value of M; if not, explain the reason
There is no existence. From the meaning of the question, we can get: M2 ≠ 0; therefore, m ≠ 0, and △ = [- (2m-1)] 2-4m2 ≥ 0, the solution is: m ≤ 14; and M is required to be a non negative integer, so such m does not exist
1-2 + Y / 6 = y-1-2y / 4
1-2+6/y=y-1+2y/4
-1+6/y=y-1+y/2
y/6-1=y/2-1
y/6=y/2
Y=0
If x, y, Z are all positive real numbers and x ^ 2 + y ^ 2 + Z ^ 2 = 1, then the minimum value of S = (Z + 1) ^ 2 / 2xyz is
2XY ≤ x ^ 2 + y ^ 2 = 1 - Z ^ 2, only if x = y
∴S = (z + 1)^2 / 2xyz
≥ (z + 1)^2 / z(1 - z^2)
= (z + 1) / z(1 - z)
= - 1 / [(z+1) - 3 + 2/(z+1)]
Since (Z + 1) + 2 / (Z + 1) - 3 ≥ 2 √ 2 - 3, the equal sign holds when Z + 1 = 2 / (Z + 1), that is, z = √ 2-1
So - 1 / [(Z + 1) - 3 + 2 / (Z + 1)] ≥ 1 / (3 - 2 √ 2) = 3 + 2 √ 2,
The minimum value of S is 3 + 2 √ 2, which holds when z = √ 2-1
It is not difficult to find out that x = y = √ (√ 2 - 1)
It is known that the polynomial-2 ^ 6x ^ 2Y ^ m + 1-3x ^ 4Y + 1 / 3 XY ^ 3-9 is a quaternion of degree six, and the degree of the monomial 2x ^ 2ny ^ 2
The degree of this polynomial is the same, so we can find the value of 3M + 2n
It is known that the polynomial-2 ^ 6x ^ 2Y ^ m + 1-3x ^ 4Y + 1 / 3 XY ^ 3-9 is a quaternion of degree six, and the degree of the monomial 2x ^ 2ny ^ 2 is the same, so the value of 3M + 2n can be obtained
m=3,n=2
3m+2n=13
Help me to do a problem to prove: no matter what the value of M is, there are always two equal real roots for the quadratic equation x2 + (4m + 1) x + 2m-1 = 0 with respect to X
△=(4m+1)²-4(2m-1)
=16m²+8m+1-8m+4
=16m²+5>0
So there should be two unequal real roots
Solving equation 3Y ^ 2 - (2Y + 1) (Y-2) = (Y-5) (Y-1)
∵3y^2-(2y+1)(y-2)=(y-5)(y-1)
==>3y^2-2y^2+3y+2=y^2-6y+5
==>9y==3
==>y=1/3
The solution of the original equation is y = 1 / 3
Let x, y, Z be positive real numbers and satisfy X-Y + 2Z = 0, then the minimum value of y2xz is______ .
From the meaning of the question, y = x + 2Z, ∵ x, y, Z are positive real numbers, ∵ y = x + 2Z ≥ 22xz, ∵ Y2 ≥ 8xz, ∵ y2xz has a minimum value of 8, so the answer is 8
ZY ^ 2 + 3x = 1 find the value of polynomial 4Y ^ 2 + 6x-7
Because 2Y ^ 2 + 3x = 1, so 4Y ^ 2 + 6x = 2, so 4Y ^ 2 + 6x-7 = 2-7 = - 5