Finding monotone interval of function y = log 0.5 (4x-x ^ 2)

Finding monotone interval of function y = log 0.5 (4x-x ^ 2)

To form exponential function:
4x-x^2=y^0.5=1/y^2
-(x-2)^2+4=1/y^2
On the left side of the equation, the maximum value of (X-2) ^ 2 + 4 is 4 when x = 2
But 0
Finding monotone interval of function y = log with base 2 (2x ^ 2-5x-3)
Want to ask! Do you need to consider the range? For example, (2x ^ 2-5x-3) is greater than 0! But although the textbook answers list the range! But when the final interval is calculated, it ignores the range and uses 5 / 4 directly! Is there a problem with the textbook!
If you want to solve this problem, you need to find the definition field first, that is, 2x ^ 2-5x-3 > 0, that is, x > 3 or X3 time increment, X
The function defined on R is y = f (x), f (0) ≠ 0, when x > 0, f (x) > 1, and for any a, B ∈ R, f (a + b) = f (a) f (b)
(1) Verification: F (0) = 1
(2) Proof: for any x ∈ R, f (x) > 0
(3) Prove that f (x) is an increasing function on R
(4) If f (x) f (2x-x ^ 2) > 1, find the value range of X
(1) Let B = 0, then f (a) = f (a) f (0) f (0) = 1
(2) Let a = x > 0, B = - X1 > 0, so f (- x) > 0
So for any x ∈ R, f (x) > 0
(3) Let X1 > X2, X1 = x2 + m, M = x1-x2 > 0, then f (m) > 1
So f (x1) = f (x2 + m) = f (x2) * f (m) > F (x2)
So f (x) is an increasing function on R
(4) f(2x-x^2)>1
Because f (x) is an increasing function, f (0) = 1
So 2x-x ^ 2 > 0 x (X-2)
(1) Because f (a + b) = f (a) f (b), so f (0 + 0) = f (0) f (0), that is, the square solution of F (0) = f (0), because f (0) ≠ 0, then f (0) = 1
(2) Because f (x-x) = f (x) f (- x), that is 1 = f (x) f (- x), and because f (x) > 1 when x > 0, f (x) > 0 for any x ∈ R
(3) Let x2 > x1, then f (x2-x1) = f (x2) f (- x1)... Expansion
(1) Because f (a + b) = f (a) f (b), so f (0 + 0) = f (0) f (0), that is, the square solution of F (0) = f (0), because f (0) ≠ 0, then f (0) = 1
(2) Because f (x-x) = f (x) f (- x), that is 1 = f (x) f (- x), and because f (x) > 1 when x > 0, f (x) > 0 for any x ∈ R
(3) Let x2 > x1, then f (x2-x1) = f (x2) f (- x1), because in (2) the conclusion 1 = f (x) f (- x), then the original formula is f (x2-x1) = f (x2) / F (x1). Suppose f (x) is an increasing function on R, then f (x2) > F (x1), that is f (x2) / F (x1) > 1. Because when x > 0, f (x) > 1, it shows that x2 > X1 in F (x2-x1) is correct. Explain that the hypothesis is correct
(4) Simplify f (x) f (2x-x ^ 2) > 1, that is, the solution f (3x-x ^ 2) > F (0). Because f (x) is an increasing function on R, then the proposition is transformed into 2x-x ^ 2 > 0. The solution is: 0A, and because f (a + b) = f (a) f (b), and when b > 0, f (b) > 1, so f (a + b) > F (a), so f (x) is an increasing function,
④ The original formula is f (x + 2x-x ^ 2) > 1, and because it is a simple increasing function, so we get 3x-x ^ 2 > 0 from the topic, and then we solve something by ourselves, which is simple.
Think about it for yourself. I haven't come up with it yet. ... unfold
① Let a = b = 0, then F0) = 1
② Second, I can't afford it. I'm sorry
③ Let b > 0, then a + b > A, and because f (a + b) = f (a) f (b), and when b > 0, f (b) > 1, so f (a + b) > F (a), so f (x) is an increasing function,
④ The original formula is f (x + 2x-x ^ 2) > 1, and because it is a simple increasing function, so we get 3x-x ^ 2 > 0 from the topic, and then we solve something by ourselves, which is simple.
Think about it for yourself. I haven't come up with it yet. Put it away
The monotone increasing interval of the function y = log2 [1 / 2-cos (2x + π / 4)] on (0, π) is?
2 after log is the base,
The original function can be divided into y = log2 (T)
t=1/2-cos(2x+π/4)
The proposition requires that the original function monotone increase, and the function y (T) monotone increase, so t (x) monotone increase,
Cos (2x + π / 4) monotone decreasing
2kπ≤2x+π/4≤π+2kπ
kπ-π/8≤x≤3π/8+kπ
Because 0
y=log₂[1/2-cos(2x+π/4)]
Let's look at the domain first
From 1 / 2-cos (2x + π / 4) > 0
We get: cos (2x + π / 4)
If the function f (x) = log2a (x + 1) defined in the interval (0,1) satisfies f (x) > 0, then the value range of real number a
If the function f (x) = log2a (x + 1) in the interval (0,1) satisfies f (x) > 0, then the value range of real number a
In the interval (1,2), f (x) = log2a (x + 1) > 0,2a > 1 = = = > a > 1 / 2/
The range of function y = Log1 / 3 (5 + 4x-x2),
5+4x-x2=-(x-2)^2+1≥1
log1/3(5+4x-x2)≤0
One
5+4x-x2=-(x-2)^2+1≥1
log1/3(5+4x-x2)≤0
If the function f (x) = log2a (x + 1) defined in the interval (- 1,0) satisfies f (x) > 0, then the value range of a is______ .
The value range of ∵ x ∈ (- 1,0), ∵ 0 ＜ x + 1 ＜ 1. And ∵ f (x) ＞ 0, ∵ 0 ＜ 2A ＜ 1, ∵ A is (0,12)
If f (x) = ax ^ 3 + BX ^ 2-3x, a and B belong to r.1
Given the function f (x) = ax ^ 3 + BX ^ 2-3x, a and B belong to r.1. If f (x) is a monotone function on R, find the relationship between a and B. 2. When a = 1 and B = 4, find the monotone interval of F (x)
(1) The derivative of F is 3ax ^ 2 + 2bx-3, which is negative at x = 0, so the derivative of F is not positive, and a is obtained
Given that the function f (x) = 1 / 2x ^ 2-x + 3 / 2 defined on [1, M] is also [1, M], then the value range of real number m is also [1, M]
f(x)=(1/2)(x-1)^2+1
When x = 1, X belongs to [1, M], the image is on the right side of the symmetry axis. The function increases monotonically on [1, M]. When x = m, the function reaches the maximum, then
(1/2)(m-1)^2+1=m
(1/2)(m-1)^2=m-1
(1/2)(m-1)=1
m-1=2 m=3
The derivative function f (x) = - 1 / 3x ^ 3 + ax ^ 2 + BX increases monotonically in the interval [- 1,2]
1. Find the conditions of Changshu a, B and the region of point (a, b);
2. If f (x) decreases monotonically in the interval (- ∞, - 1], [2, + ∞), find the values of constants A and B
For the convenience of writing, the derivative of function f (x) is g (x), and min is the minimum
1. According to the derivation rule, G (x) = - X & sup2; + 2aX + B symmetry axis line x = a,
∵ f (x) increases monotonically in [- 1,2]
That is, G (x) (min) ≥ 0
① When - 1 < a < 2, because g (- 1) = b-1-2a, G (2) = B + 4a-4
When 1 / 2 < a < 2, the minimum value of G (x) is b-1-2a ≥ 0, that is, B ≥ 1 + 2A
When - 1 < a ≤ 1 / 2, the minimum value of G (x) is B + 4a-4 ≥ 0, that is, B ≥ 4-4a
② When a ≥ 2, G (x) increases monotonically in [- 1,2], so B ≥ 1 + 2A
③ When a ≤ - 1, G (x) decreases monotonically in [- 1,2], so B ≥ 4-4a
When a > 1 / 2, a and B satisfy the condition that B ≥ 1 + 2A
When a ≤ 1 / 2, a and B satisfy the condition that B ≥ 4-4a
I can draw the area myself
（2）g(-1）=0 g（2）=0
We can get a, B
F (x) = - 1 / 3x ^ 3 + ax ^ 2 + BX increases monotonically in the interval [- 1, 2]
The derivation shows that f '(x) = - x ^ 2 + 2aX + B is greater than 0 in the interval [- 1, 2]
X ^ 2-2ax-b is less than 0 in the interval [- 1, 2]
Let g (x) = x ^ 2-2ax-b
g(-1)