Let f (x) = X3 + bx2 + CX, G (x) = f (x) - F ′ (x). If G (x) is an odd function, find the value of B and C

Let f (x) = X3 + bx2 + CX, G (x) = f (x) - F ′ (x). If G (x) is an odd function, find the value of B and C

From F (x) = X3 + bx2 + CX, f ′ (x) = 3x2 + 2bx + C, then G (x) = f (x) - F ′ (x) = X3 + (B-3) x2 + (c-2b) x-C, ∵ g (x) is an odd function, ∵ G (0) = - C = 0, C = 0. ∵ g (x) = X3 + (B-3) x2-2bx

Given the function f (x) = X3 + bx2 + CX + D, take the extremum 2 (1) at x = 0 to find the value of C and D (2) try to study the number of tangent lines of the curve y = f (x) perpendicular to the straight line x-by + 1 = 0 (3) if any x belongs to [1,2], there exists t belonging to (0,1), such that et-1nt-1

(1) F (0) = 2, so d = 2 and x = 0 take the extremum, so f '(0) = 0 = C, so C = 0 (2) the slope of straight line is 1 / B, the slope of curve tangent is 3x ^ 2 + 2bx, the vertical condition is (3x ^ 2 + 2bx) / b = - 1, when 4B ^ 2-12b > 0, that is, b > 3, there are two tangents, when B = 3, there is one tangent, when B = (...)

A = {x | 1 ≤ x ≤ 4} f (x) = x ^ 2 + PX + Q and G (x) = 4 / x + X are functions defined on a
The maximum value of F (x) on a is obtained by taking the minimum value at x 0 and satisfying f (x 0) = g (x 0)

When x ＞ 0, x + 4 / X ≥ 2 radical (x × 4 / x) = 2 × 2 = 4. If and only if x = 4 / x, i.e. x = 2, the inequality takes the equal sign. Because 2 is on the interval [1,4], so x0 = 2, G (x0) = f (x0) = 4, because f (2) is the minimum value on the interval [1,4], so x = 2 is the symmetry axis of F (x), i.e. - P / 2 = 2 → P = - 4f (x) = x & sup2

Given that the functions f (x) = x2 + PX + Q and G (x) = x + 4x are defined on a {x | 1 ≤ x ≤ 52}, for any x ∈ a, there exists a constant x0 ∈ a such that f (x) ≥ f (x0), G (x) ≥ g (x0), and f (x0) = g (x0), then the maximum value of F (x) on a is ()
A. 52B. 174C. 5D. 4140

From the known functions f (x) = x2 + PX + Q and G (x) = x + 4x have the minimum values f (x0), G (x0) in the interval [1,52], and because the minimum value of G (x) = x + 4x & nbsp; in the interval [1,52] is g (2) = 4, f (x) min = f (2) = g (2) = 4, we get: − P2 = 24 + 2p + q = 4, that is, P = − 4q = 8, so we get: F (x) = x2-4x + 8 ≤ f (1) = 5

If the formula of quadratic function y = x square-2x-3 is y = a (x = m), then y=

First, find the vertex of the function as (1, - 4) and write it as the square of the vertex formula y = (x - 1) - 4

Find the formula form of y = 1 / 2x ^ 2-6x + 21
I ended up with y = (X-6) ^ 2 + 6. What's going on

y=1/2(x^2-12x)+21
=1/2(x^2-12x+36-36)+21
=1/2(x^2-12x+36)-18+21
=1/2(x-6)^2+3

(2010, Anhui) if the quadratic function y = x2 + BX + 5 is y = (X-2) 2 + K, then the values of B and K are ()
A. 0，5B. 0，1C. -4，5D. -4，1

∵ y = (X-2) 2 + k = x2-4x + 4 + k = x2-4x + (4 + k), and ∵ y = x2 + BX + 5, ∵ x2-4x + (4 + k) = x2 + BX + 5, ∵ B = - 4, k = 1

How to formulate quadratic function? For example, y = the square of X + X + 3,

y=x²+x+3=x²+2×1/2×x+1/4-1/4+3
=(x+1/2)²+11/3

Find the coordinates of the intersection of the image of the following quadratic function and the Y axis (1) y = x ^ 2 + 6x + 9 (2) y = 9-4X ^ 2 (3) y = (x + 1) ^ 2-9

(1)y=x^2+6x+9
When x = 0, y = 9
The coordinates of the intersection of the image of quadratic function and Y axis are (0,9)
(2)y=9-4x^2
When x = 0, y = 9
The coordinates of the intersection of the image of quadratic function and Y axis are (0,9)
(3)y=(x+1)^2-9
When x = 0, y = - 8
The coordinates of the intersection of the image of quadratic function and Y axis are (0, - 8)

It is known that the vertex of quadratic function image is (- 1,2) and passes through the point (1, - 3)
2. It is known that the image of quadratic function passes through points (4,3), and when x = 3, y has a maximum value of 4

1.y=ax^2+bx+c
2=a-b+c
-3=a+b+c
-b/2a=-1
Solution
b=-5/2
a=-5/4
c=3/4
∴y=-5/4x^2-5/2x+3/4
2.y=ax^2+bx+c
-b/2a=3
3=16a+4b+c
4=9a+3b+c
a=-1
b=6
c=-5
∴y=-x^2+6x-5
I'm glad to answer for you. I hope it will help you. If you have any questions, you can communicate with me