Given that in a function y = (1 + 2m) x, the value of function y decreases with the increase of independent variable x, then the value range of M is () A. m≤−12B. m≥−12C. m＜−12D. m＞−12

Given that in a function y = (1 + 2m) x, the value of function y decreases with the increase of independent variable x, then the value range of M is () A. m≤−12B. m≥−12C. m＜−12D. m＞−12

∵ function y = (1 + 2m) x-3 is a linear function. To make the value of function y decrease with the increase of independent variable x, ∵ 1 + 2m ＜ 0, the solution is m ＜ - 12

Given a function y = (1-2m) x + m + 1, the value of the function y decreases with the increase of the value of the independent variable x, m < 1 / 2 1) is the intersection of the image of this function and the Y axis m on the positive half axis or the negative half axis of the Y axis

The linear function y = (1-2m) x + m + 1, and the function value y decreases with the increase of the independent variable x
So 1-2m1 / 2
m+1>3/2
So it intersects the y-axis with the positive half axis

How to find the monotone interval of function f (x) = 2x ^ 3-3x ^ 2-12x + 13 by list

First, the derivative function is f '= 6x ^ 2-6x-12 = 6 (x ^ 2-x-2) = 6 (X-2) (x + 1) for listing (- infinity, - 1) - 1 (- 1,2) 2 (2, + infinity) f' greater than 0 equal to 0 less than 0 equal to 0 greater than 0, so the monotone increasing interval is (- infinity

If the minimum value of the function f (x) = - 2x ^ 3-3x ^ 2 + 12x + 1 in the interval [M, 1] is - 17, then M =?
Arena 19:51:58
f(x1,x2,x3) = x1^2 + x2^2 + x3^2 - 2*x1*x2 -2*x2*x3 + 2*x1*x3
Using a symmetric matrix A, the function f is expressed in the form of matrix

∵ f '(x) = - 6x & sup2; - 6x + 12, Let f' (x) = 0, then the stationary point x = - 2, x = 1
When x

The function f (x) = - 2x ^ 3-3x ^ 2 + 12x + 1 in the interval [M, 1] (1) if the minimum value is - 17, find the value of the real number m; (2) if the minimum value is - 19, find the value range of the number M

F (x) = - 2x ^ 3-3x ^ 2 + 12x + 1, f '(x) = - 6 (x + 2) (x-1) = = = > F' (- 2) = f '(1) = 0. = = = > function f (x) decreases on (- ∞, - 2] and increases on [- 2,1], and f (- 2) = - 19, f (1) = 8. (1)

Given that the function y = LG (MX ^ 2-4mx + m + 3) is meaningful, find the value range of real number m satisfying the following conditions: (1) any x belongs to R; (2) Any Y belongs to R

If M is not equal to 0, the maximum value of MX ^ 2-4mx + m + 3 must be greater than 0, so Δ = (4m) & sup2; - 4m (M + 3) < 0 and M > 0, so the range is [0,1) 2. M = 0. The maximum value of MX ^ 2-4mx + m + 3 must be positive and infinite, and the minimum value must not be greater than 0, so Δ = (4m) & sup2; - 4m (M + 3) ≥ 0 and M

If the minimum value of function f (x) = x ^ 2-2ax + 1 (a belongs to real number) on [- 1,1] is - 3, find the value of A

Let f (- 1) = 1 + 2A + 1 = - 3A = - 5 / 2F (x) = x ^ 2 + 5x + 1F (1) = 7F (1) > F (- 1) Let f (1) = 1-2a + 1 = - 3A = 5 / 2F (x) = x ^ 2 + 5x + 1F (1) = 7F (1) > F (- 1) Let f (1) = 1-2a + 1 = - 3A = 5 / 2F (x) = x ^ 2-5x + 1F (- 1) = 7a = - 5 / 2 F (- 1) = - 3 be the minimum value a = 5 / 2

Given the function f (x) = 1 / 3x3-1 / 2x2 ax, and f (x) obtains the extremum at x = 2 (1) find the monotone interval of F (x) (2) find the maximum and minimum of F (x) on X? [0,3]

Derivation:
f'(x)=x(sqrt)2-x-a
f'(2)=4-2-a=0
=> a=2
=>f'(x)=(x-2)(x+1)
=> -1

The function f (x) = MX ^ 3 NX ^ 2 is known. When x = 1, f (x) has a maximum. 2. Find the value of M and N? Find the minimum of F (x)?
Look at the title,

The derivative f '(x) = 3mx & sup2; - 2nx, where f' (1) = 0 and f (1) = 2, is obtained

The maximum value of F (x) = - x ^ 2 + MX + 1 in the interval [- 2, - 1] is the maximum value of function f (x), then the value range of M is

The solution shows that the axis of symmetry x = - B / 2A = - M / 2 * (- 1) = m / 2 belongs to the interval [- 2, - 1]
That is - 2 ≤ M / 2 ≤ - 1
That is - 4 ≤ m ≤ - 2