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

Let the function f (x) = X3 + bx2 + CX, G (x) = f (x) - F ′ (x). If G (x) is an odd function, find the values 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. From G (- 1) = - 1 + B-3 + 2B = 3b-4, - G (1) = - 1-B + 3 + 2

Let the function f (x) = x ^ 3 + BX ^ 2 + CX (x ∈ R), and let g (x) = f (x) - f ` (x) be an odd function. Find B, C

f'(x)=3x^2+2bx+c
So g (x) = x ^ 3 + (B-3) x ^ 2 + (c-2b) x-c
g(-x)=-x^3+(b-3)x^2-(c-2b)x-c
Is an odd function
g(-x)=-g(x)
-x^3+(b-3)x^2-(c-2b)x-c=-x^3-(b-3)x^2-(c-2b)x+c
2(b-3)x^2-2c=0
b-3=0,c=0
b=3,c=0

The function f (x) = - 1 / 3x ^ 3 + BX ^ 2 + CX + BC about X is known, and its derivative is f '(x). Let g (x) = lf' (x) l, The function f (x) = - 1 / 3x ^ 3 + BX ^ 2 + CX + BC is known, and its derivative function is f '(x). Let g (x) = | f' (x) |, and record that the maximum value of function g (x) in interval [- 1,1] is m (1) If the function f (x) has a limit value of - 4 / 3 at x = 1, try to determine the values of B and C, (2) If | B | > 1, it is proved that there is m > 2 for any C (3) If M > = k is constant for any B, C, try to find the maximum value of K My first question is how to calculate B = 1 and C = - 1, right?

Let me try... After sleeping... (1) f (x) = - 1 / 3x ^ 3 + BX ^ 2 + CX + BCF '(x) = - x ²+ 2bx + C by question, f '(1) = - 1 + 2B + C = 0f (1) = - 1 / 3 + B + C + BC = - 4 / 3bC + B + C + 1 = (B + 1) (c + 1) = 0, B = - 1 or C = - 1 if B = - 1, - 1-2 + C = 0, C = 3 if C = - 1, - 1 + 2b-1 = 0, B = 1, so (B, c) = - 1,3) or (1

The monotonicity, convexity, extremum and inflection point of function y = x cube + 3x square - 9x + 1 are discussed in the list

Derivative
y'=3x^2+6x-9
Judge this quadratic function
When y 'is greater than 0, it increases monotonically; when y' is less than 0, it decreases monotonically, and the range of X is easy to find
Re derivation
y''=6x+6
When y '' is greater than 0, it is a convex function, and when y '' is less than 0, it is a concave function, and the range of X is easy to find
In addition, y '= 0, find x, bring it into the original formula, and find y is the extreme value
And y '' = 0, and finding X and the corresponding y is the inflection point

Monotone interval and extreme value of function y = 1 / 3x ^ 3-2x ^ 2 + 5, please answer it. Thank you very much!

Y '= x ^ 2-4x = x (x-4) = 0, x = 0,4
Monotonic increasing interval: X4
Monotone decreasing interval: 0

It is known that x = 3 is an extreme point of the function f (x) = 1 / 3x ^ 3-ax ^ 2 + 3x. Find a? Find the monotone interval and extreme value of the function? Please, gods Hurry!

F '(x) = x ^ 2-2ax + 3 F' (3) = 9-6a + 3 = 0 a = 2 F (x) = 1 / 3x ^ 3-2x ^ 2 + 3x f '(x) = x ^ 2-4x + 3 Let f' (x) = 0 get x = 1, X = 3 ‰ f (x) increase on (- ∞, 1), decrease on (1,3), and increase on (3, + ∞): F (1) = 1 / 3-2 + 3 = 4 / 3 F (3) = 9-18 + 9 = 0