Given the real number m > 0, the function f (x) = MX (X-2) ^ 2 has a maximum value 8. Find the value of real number M

Given the real number m > 0, the function f (x) = MX (X-2) ^ 2 has a maximum value 8. Find the value of real number M

f(x)=mx³-4mx²+4mx
f'(x)=3mx²-8mx+4m=m(3x-2)(x-2)
Let f '(x) = 0. Then x = 2 / 3 or x = 2
Because m > 0
So when x = 2 / 3, we take the maximum
f(2/3)=2m/3×16/9=8
The solution is m = 27 / 4
If f '(x) = m (3x-2) (X-2) and f' (x) = 0, then x = 2 / 3 and 2, we can see that there is a maximum at 2 / 3 and a minimum at 2. Substituting f (x), we can get m = 27 / 4
If the function f (x) = x2 + ax holds f (1 + x) = f (1-x) for any real number x, find the value of real number & nbsp; a
For any real number x, f (1 + x) = f (1-x) holds for ∵ function f (x) = x2 + ax, ∵ axis of symmetry x = - A2 = 1, ∵ a = - 2,
If there are exactly three integers in the solution set where the square of X inequality (2x-1) is less than the square of a (x), then the value range of real number a is?
If there are exactly three integers in the solution set where the square of X inequality (2x-1) is less than the square of a (x), then the value range of real number a is?
How do you do
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Let f (x) = (2x-1) ^ 2-ax ^ 2
=There are exactly three integers of (4-A) x ^ 2-4x + 1 > 0
So 4-a0
Let f (x) = 0, X1 + x2 = 4 / (4-A), X1 * x2 = 1 / (4-A)
lx1-x2l^2=(x1+x2)^2-4x1x2
Because of the integer solution, so 2
The axis of symmetry of F (a + x) = f (b-X) and the axis of symmetry of function y = f (a + x) and function y = f (b-X)
Notice! It's from the book. Their axes of symmetry are x = (a + b) / 2; X = (B-A) / 2. I want to know how it comes out
First: the axis of symmetry of F (a + x) = f (b-X) is x = (a + b) / 2. Note that this is an axisymmetric function image, which is an image. First of all, we need to know a relationship: if f (a + x) = f (A-X), then x = a symmetry, and we can infer that if f (x) = f (2a-x), then x = a symmetry, so I
Given that the constant T is a negative real number, what is the domain of the function f (x) = √ 12t ^ 2-tx-x ^ 2
First: 12t ^ 2-tx-x ^ 2 > = 0
Then the factor is decomposed into:
（3t-x)(4t+x) >=0
Then 3t-x > = 0 and 4T + x > = 0 get x = 4T
Or 3t-x
If the function y = f (x) has f (3 + x) = f (1-x) for all real numbers x, then the image of y = f (x) has an axis of symmetry () if the function y = f (x) has f (3 + X%) for all real numbers X
How to choose and fill in the blanks:
3+1/2=2
The axis of symmetry is a straight line x = 2
When learning periodic function, we should talk about this method
How to solve the problem:
Let x = 1, then f (4) = f (0),
Let x = 0, then f (3) = f (1),
Let x = - 1, then f (2) = f (2)
Therefore, through the law, the straight line x = 2 is the axis of symmetry
Because f (3 + x) = f (1-x)
And 3 + X - (1-x) = 2 * x + 2
So the function is symmetric with respect to y = 2 * x + 2
I didn't understand the question in the back...
Given that the constant T is a negative real number, what is the domain of definition of the function f (x) = √ 12t-tx-x? The answer is [3T, - 4T],
It should be f (x) = √ 12t ^ 2-tx-x ^ 2, 12t-tx-x ^ 2 ≥ 0 x ^ 2 + tx-12t ^ 2 ≤ 0 (x-3t) (x + 4T) ≤ 0
T
Decompose the factor in the root sign into - (x-3t) * (x + 4T), because the value under the root sign needs to be greater than or equal to zero, so the formula I wrote should be greater than or equal to zero, that is, (x-3t) * (x + 4T) is less than or equal to zero, so the solution of X is as above.
12t-tx-x ≥ 0?: can you be more detailed?
If the function f (x) = 2cos (ψ x + φ) + m has a straight line x = π / 8 and f (π / 8) = - 1, then the value of real number m is equal to___
For the function cosy = x,
Its axis of symmetry is the x value corresponding to the extreme value of the function!
That is to say, when x is on the axis of symmetry, the function y = 1 or - 1
Therefore:
2+m=-1
Or:
-2+m=-1
The solution is as follows
M = - 3 or 1
Given that the constant T is a negative real number, the monotone increasing interval of the function f (x) = - √ 12t ^ 2-tx-x ^ 2 is
The derivative function is equal to - t-2x so that it is greater than 0 X
Let f be a function from real number set R to real number set R, satisfying that f (x + y) = f (x) + F (y) + 2XY, if the image of F (x) has symmetry axis X = k, and in the interval [2,3]
Let f be a function from the real number set R to the real number set R, satisfying that f (x + y) = f (x) + F (y) + 2XY. If the image of F (x) has a symmetry axis X = K and decreases monotonically in the interval [2,3], the value range of K is obtained
How to find f (1)
F (K + 0) = f (k) + F (0) + 0, f (0) = 0;
f(x-x)=f(x)+f(-x)-2x^2=0,f(x)+f(-x)=2x^2;
If f (K + x) = f (K-X), f (- x) - f (x) = 4kx can be obtained;
By subtracting the two formulas, f (x) = x ^ 2-2kx;
The square opening of F (x) = (x-k) is upward
Monotonically decreasing from 2 to 3 indicates that the axis of symmetry K is greater than or equal to 3
Let y = 1, then f (x + 1) = f (x) + F (1) + 2x,
f(x+1)-f(x)=f（1）+2x
f(x)-f(x-1)=f（1）+2（x-1)
。。。。
f(2)-f(1)=f(1)+2*1
F (x + 1) - f (1) = f (1) x + 2 (1 + 2 +)... X)
f(x+1)=x^2+[1+f(1)]x+f(1)
f(x)= x^2+[f(1)-1]x
Only when k > = 3 can it satisfy the problem
To sum up, k > = 3
V