Given that f (x) is an odd function on R, and when x > 0, f (x) = 3x ^ 3 + 2x ^ 2-1, find the analytic expression of F (x)

Given that f (x) is an odd function on R, and when x > 0, f (x) = 3x ^ 3 + 2x ^ 2-1, find the analytic expression of F (x)

Take x ＜ 0, then - x ＞ 0. Bring it into the original function, get: F (- x) = - 3x ^ 3 + 2x ^ 2-1, because f (x) is odd function, so f (- x) = - f (x), deduce f (x) = 3x ^ 3-2x ^ 2 + 1
Let f (x) = x2 + BX + C, & nbsp; & nbsp; X ≤ 02, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; x > 0, if f (- 4) = f (0), f (- 2) = - 2, then the number of zeros of G (x) = f (x) - x is zero______ .
From F (- 4) = f (0), we can get 16-4b + C = C, and the solution is b = 4. From F (- 2) = - 2, that is, 4-8 + C = - 2, we can get C = 2. So f (x) = x2 + 4X + 2, X ≤ 02, x > 0, from G (x) = 0, we can get f (x) = X. in the same coordinate system, we can make the function y = f (x), y = x image respectively, as shown in the figure: from the image, we can know two images
Given that the function f (x) is an odd function and its domain of definition is r, when x < 0, f (x) = 2x ^ 2-3x + 1, then the analytic expression of F (x) is?
f(x)=2x2+3x+1 (x>0)
f(0)=0 (x=0)
Let f (x) = 2x ^ 2-3x + 1 in the part of x > 0 be symmetrical about the origin, let x0,
f(x)=-f(-x)=-[2(-x)^2-3(-x)+1]=-[2x^2+3x+1]=-2x^2-3x-1;
When x = 0, it is discontinuous, x0, f (x) = - 1; x > 0, X -- > 0, f (x) = 1; in order to define the function when x = 0, we can define f (x) = (- 1 + 1) / 2 = 0 when x = 0;
If the function f (x) = x & sup2; - ax + B has two zeros 2 and 3, try to find the zeros of G (x) = BX & sup2; - ax + 1
Because 2 is the zero of F (x), so
2 & # 178; - 2A + B = 0, divide both sides by 2 & # 178; to get:
(1/2)²b-a(1/2)+1=0
So 1 / 2 is a zero point of G (x), similarly 1 / 3 is also a zero point of G (x);
So the zero point of G (x) has
1/2,1/3
The coefficients are arranged in reverse, and the root is the reciprocal of the root of the original equation.
So the zeros of G (x) are 1 / 2, 1 / 3
It is proved that on the interval [2,5], the function f (x) = - 2x, square = 3x - 1 is reduced
f(x)=-2x^2+3x-1=-2(x^2-3x/2)-1=-2(x-3/4)^2-1+9/8=-2(x-3/4)^2+1/8
It can be seen that the vertex of the parabola is (3 / 4,1 / 8), and the opening is downward. Therefore, it can be seen that when x > 3 / 4 on the right side of the vertex, the function is a decreasing function, x > 3 / 4
If the two zeros of function f (x) = x2-ax-b are 2 and 3, then the zeros of function g (x) = bx2-ax-1 are 2 and 3______ .
From the meaning of the question: 4-2a-b = 09-3a-b = 0, the solution of a = 5B = - 6  g (x) = - 6x2-5x-1 has zero points of - 12, - 13. So the answer is: - 12, - 13
If f ((1-2x) / (3 + x)) = 3x-1, find f (x)
The problem of finding the content of function domain
Let t = (1-2x) / (3 + x), then x = (1-3T) / (T + 2), so f (T) = 3x-1 = (1-10T) / (T + 2), so f (x) = (1-10x) / (x + 2), the domain of definition is x ≠ - 2
Let 1-2x / 3 + x = t, 1-2x = 3T + TX, (2 + T) x = 1-3T x = (1-3T) / (2 + T) f (T) = 3 * (1-3T) / (2 + T) - 1 f (x) = (3-9x) / (2 + x) - 1 = (1-10x) / (2 + x)
If the zero point of function f (x) = ax + B is 2, then the zero point of function g (x) = bx2 ax is ()
A. 0，2B. 0，12C. 0，-12D. 2，12
∵ function f (x) = ax + B has a zero point of 2, ∵ 2A + B = 0, {B = - 2A, ∵ g (x) = bx2 AX = - 2ax2 AX = - ax (2x + 1), ∵ ax (2x + 1) = 0 {x = 0, x = - 12 ∵ function g (x) = bx2 ax has zero points of 0, - 12
If f (2x + 1) = 3x-2 and f (a) = 4, then the value of a is______ .
Let t = 2x + 1, x = t − 12, substitute f (2x + 1) = 3x-2, f (T) = 32t − 72, then f (x) = 32x − 72, then f (a) = 32a − 72 = 4, the solution is a = 5, so the answer is: 5
Given that the two zeros of function f (x) = x2-ax-b are 2 and 3, then the zeros of function g (x) = bx2-ax-1 are ()
A. - 1 and - 2b. 1 and 2C. − 12 and − 13D. 12 and 13
The two zeros of ∵ function f (x) = x2-ax-b are 2 and 3 ∵ the two real roots of equation x2-ax-b = 0 are 2 and 3. According to Weida's theorem, the two zeros of ∵ function f (x) = x2-ax-b are − 12 and − 13, that is, function g (x) = - 6x2-5x-1 ∵ - 6x2-5x-1 = 0 ⇔ (2x + 1) (3x + 1) = 0 ∵ g (x) = 0