Given that the domain of function f (x) is r, f (x) = 2x-x2 when f (x) = - f (- x) and x > 0;

Given that the domain of function f (x) is r, f (x) = 2x-x2 when f (x) = - f (- x) and x > 0;

x=0
f(x)=-f(-x)
f(0)=-f(0)
f(0)=0
x0
So f (- x) = 2 (- x) - (- x) & # 178; = - 2x-x & # 178;
So f (x) = - f (- x) = 2x + X & # 178;
So f (x)=
2x+x²,x0

It is known that y = f (x) is an odd function of R in the domain of definition. When x ≥ 0, f (x) = x2-2x. Why is the answer to the analytic expression of F (x) on r not f (x) = x ^ 2 + 2x but f (x) = - x ^ 2-2x

For odd functions
f(x) = - f(-x)
So when x ≥ 0, f (x) = x2-2x
So f (- x) = - f (x) = - x2 + 2x
So let t = - x, so t

If x is less than 0, f (x) = - x2 + 2x + 1, the analytic expression of F (x)
Is when x is greater than 0

x0;f(x)=x2+2x-1
x=0;f(x)=0

It is known that y = f (x) is an odd function whose domain is R. when x ≥ 0, f (x) = x2-2x. Then the analytic expression of F (x) on x < 0 is ()
A. f（x）=x2+2xB. f（x）=-x2+2xC. f（x）=x2-2xD. f（x）=-x2-2x

When x ＜ 0, - x ＞ 0, f (- x) = (- x) 2-2 (- x) = x2 + 2x, and f (x) is an odd function, so f (x) = - f (- x) = - x2-2x, so D

If x belongs to [0,2], f (x) = - x ^ 2 + 2x-1, find f (x)

If f (x) is an even function, then f (x) = f (- x)
When 0

Given that the image of the function f (x) = ax ^ 2 + BX (a ≠ 0) is symmetric with respect to the line x = 1, and that the equation f (x) = x has equal real roots, the analytic expression of F (x) is obtained

The image is symmetric with respect to the line x = 1, so - B / (2a) = 1, B = - 2A
F (x) = x has equal real roots, that is, ax ^ 2 + (B-1) x = 0 has equal real roots, so the discriminant = (B-1) ^ 2 = 0, B = 1
So a = - 1 / 2
f(x)=-1/2*x^2+x

If the two roots of the equation AX & # 178; + BX + B = 0 (a < 0) are x1x2 and satisfy X1 < X2, then the solution of the inequality ax & # 178; + BX + B < 0 is_____ .

ax²+bx+b=a(x-x1)(x-x2)
a(x-x1)(x-x2)0
So x > X2, or X

Let the eccentricity e of the ellipse x2a2 + y2b2 = 1 (a > 0, b > 0) be 12, the right focus f (C, 0), and the two roots of the equation AX2 + bx-c = 0 be X1 and X2 respectively, then the point P (x1, x2) is in ()
A. In circle x2 + y2 = 2, B. in circle x2 + y2 = 2, C. in circle x2 + y2 = 2, D

∵ X1 + x2 = - BA, x1x2 = - cax12 + X22 = (x1 + x2) 2-2x1x2 = B2 + 2aca2e = CA = 12 ∵ a = 2cb2 = a2-c2 = 3c2, so X12 + X22 = 3c2 + 4c24c2 = 74 < 2, so a is selected in the circle

It is known that the ratio of two quadratic equations ax ^ 2 + BX + C = 0 (a does not = 0) with respect to X is 2:1. It is proved that 2B ^ 2 = 9ac
I'm in a hurry. If the answer is correct, I'll give you points

The formula X1 + x2 = - B / a 1 is obtained from WIDA's theorem
Formula X1 * x2 = C / a 2
So (x1 + x2) ^ 2 = X1 ^ 2 + 2x1x2 + x2 ^ 2 = B ^ 2 / A ^ 2 3
Divide formula 3 by Formula 1 to get formula 1
Formula X1 / x2 + 2 + x2 / X1 = B ^ 2 / AC 4
And because X1 / x2 = 2, substituting into 4 formula:
2+2+1/2=b^2/ac
It is concluded that 2B ^ 2 = 9ac
Get proof

Given that a, B, C ∈ R, and a < 0, 6a + B < 0, Let f (x) = ax ^ 2 + BX + C, compare the size of F (3) and f (π)

f(3)>f(π)
It can be calculated that the axis of symmetry of B / a > - 6 function is less than 3, because the opening is downward, so the function will decrease in the range of x > 3