The function f (x) is a decreasing function on [- 2,2], and f (m-1) - f (2m-1) is greater than 0

The function f (x) is a decreasing function on [- 2,2], and f (m-1) - f (2m-1) is greater than 0

From the domain, we can get - 2
Because f (m-1) > F (2m-1)
It is also a decreasing function on [- 2,2], so (m-1)
What is the range of the function y = cos square x-cosx + 5 / 4? (I don't know how to deal with y = cos square x-cosx + 5 / 4)
Let t = cosx, then y = T ^ 2-T + 5 / 4, - 1
Given that the function f (x) = 4 ＾ x + M2 ＾ x-6m has exactly one zero point, then the value range of real number m
Let t = 2 ^ x, then t > 0,
f(x)= g(t)=t^2+mt-6m ,
Since f (x) = 0 has exactly one zero point, G (T) = 0 has only one positive root,
(1) If G (T) = 0 has two equal positive roots, then the discriminant = m ^ 2 + 24m = 0, and X1 + x2 = - M > 0, X1 * x2 = - 6m > 0, the solution is m = - 24;
(2) If G (T) = 0 has exactly one positive root and one non positive root, then G (0) = - 6m0 because the opening of parabola is upward;
In conclusion, if f (x) = 4 ^ x + m * 2 ^ x-6m has exactly one zero point, then the value range of M is {m | M = - 24 or M > 0}
4^x+m2^x-6m=0
Let 2 ^ x = t, t > 0
Then 4 ^ x = T ^ 2
t^2+mt-6m=0
m^2+24m=0
m=-24, m=0
When m = 0, 4 ^ x = 0 is not satisfied.
So m = - 24
The monotonicity of function f (x) = x + √ (x ^ 2 + 1) on (- ∞, + ∞) is determined by definition
= (x-y) + x^2 - y^2/√(x^2+1)+√(y^2+1)
= (x-y)*{ 1 + x+y/√(x^2+1)+√(y^2+1)}
How is the above two steps transformed in the process of solving the problem?
The more detailed the process, the better. Is there any formula?
That is, x + √ X & sup2; + 1) - Y - √ (Y & sup2; + 1) = (X-Y) + [√ X & sup2; + 1) - √ (Y & sup2; + 1)] numerator denominator multiplied by [√ X & sup2; + 1) + √ (Y & sup2; + 1)] = (X-Y) + [√ X & sup2; + 1) - √ (Y & sup2; + 1)] [√ X & sup2; + 1] / [√ X & sup2; + 1) + √ (Y & sup2; + 1)]
If the image of the function y = (2m-1) x and y = 3 − MX have no intersection, then the range of M is______ .
According to the meaning of the question, we can get that the image of ∵ function y = (2m-1) x and y = 3 − MX has no intersection point, and ∵ function (2m-1) x = 3 − MX has no solution, that is △ = 4 (2m-1) (3-m) ＜ 0, ∵ function 2m − 1 ＞ 03 − m ＜ 0 or function 2m − 1 ＜ 03 − m ＞ 0
Given the function FX = x + 2x 1 / 2 + 2, X ∈ [1, + ∞] judge the monotonicity of FX in the interval [1, + ∞]
Given the function y = (2m + 1) x + M-3, if the function is a linear function and Y decreases with the increase of X, the value range of M is obtained
This function is a linear function, and Y decreases with the increase of X, 2m + 1 < 0, so m < - 0.5
Given the function y = - x ^ 2 + 4x-2 (1) if x ∈ [0,3], find the maximum and minimum of the function (2) if x ∈ [3,5], find the maximum and minimum of the function
1、
y=-x²+4x-4+2
=-(x-2)²+2
The opening is downward and the axis of symmetry x = 2
Zero
y=-x^2+4x-2
=-（x-2）^2 +2
Axis of symmetry x = 2
1) If x ∈ [0,3]
When x = 2, y has a maximum of 2
When x = 0, y has a minimum value of - 2
2) If x ∈ [3,5], the domain of definition is to the right of the axis of symmetry
So when x = 5, y has a minimum value of - 7
When x = 3, y has a maximum value of 1
Given the function y = (1-2m) X-2, in order to make the function value y increase with the increase of X, the value range of M is
1-2m＞0
M
M is less than half!!
According to the meaning of the title:
1-2m>0
That is to say, M0 is m
For the quadratic function y = - 4x ^ 2 + 8x-3, find the maximum or minimum value of the function, and analyze the monotonicity of the function
Opening down
There is a maximum
When x = - B / 2A = 8 / (2 * 4) = 1, there is a maximum y = 1
When x ≤ 1, it increases monotonically
x> Monotonically decreasing at 1
The one on the first floor is wrong. It's the maximum. It's not the minimum