What is the negative half power of (0.25) ∧ - 0.5 (0.25)

What is the negative half power of (0.25) ∧ - 0.5 (0.25)

Two
Wrong, it's 2
The range of function y = | x + 2 | + √ (x-3) is
Thank you·······
The original function is divided into two functions: F (x) = IX + 2 |, G (x) = √ (x-3). It is easy to know that only if x > = 3 is, G (x) is meaningful, and at this time, f (x) increases monotonically. Therefore, when x = 3, the original function obtains the minimum value = 5, and this function has no maximum value, so the range [5, positive infinity]
X + x 1 / 2 equals 3, x 1 / 2 minus x 2 / 1 equals 3,
 
Monotone decreasing interval of function y = log lower right half (x-2x)
The definition field is X & # 178; - 2x = x (X-2) > 0
So x2
Let f (x) = x & # 178; - 2x = (x-1) & # 178; - 1
Is the parabola with the opening upward, the axis of symmetry x = 1
When x ≤ 1, single minus
When x ≥ 1, single increment
Because y = log (1 / 2) (X & # 178; - 2x)
The base number is 1 / 2. When the true number increases, the function decreases
So x ≥ 1
Binding domain
The monotonic decreasing interval should be: x > 2
(2，＋∞)
x²－2x>0
Then: x > 2 or x > 2
A kind of operation "*": a * b = B power of a, such as 3 * 2 = 2 power of 3 = 9, then half * 3 = 1（
A * b = B power of a, if 3 * 2 = 2 power of 3 = 9, then half * 3 = (1 / 2) ^ 3 = 1 / 8
I'll let you know when I give you some points
(1 / 2) find the monotone interval of the function y = log half sin2x + quarter. Find the function f (x) = 2sinx square + 2sinx -- half, X belongs to sextile
On the monotone function of the quarter of (log / 2 + 2)
Find the function f (x) = 2sinx square + 2sinx -- half, X belongs to one sixth, one sixth
Y = log2x is increasing
Monotone increasing interval
It's sin (2x + π / 4) increasing
The answer is 0 + 2kpai
Find the monotone interval of the function y = log half sin2x + quarter.
Find the function f (x) = 2sinx square + 2sinx -- half, X belongs to one sixth, one sixth
The first floor is great!!
Given the value of F (x) = (M & # 178; - m-1) x ^ - 5m-3. M, f (x) is a power function
2. It is a power function and (0. + no group) 3. It is a positive proportion function 4. It is an inverse proportion function 5. It is a quadratic function
It's f (x) = (M & # 178; - m-1) x ^ (- 5m-3)!
1. If f (x) is a power function, then M & # 178; - M-1 = 1, the solution is m = - 1 or M = 2
2. Is a power function and the domain of definition is (0, + ∞)!
If f (x) is a power function, then M = - 1 or M = 2
When m = - 1, f (x) = x & # 178;, the domain of definition is r, incompatible;
When m = 2, f (x) = x ^ (- 13), the domain of definition is {x ∈ R | x ≠ 0 |}
So m, which is a power function and the domain of definition is (0, + ∞), does not exist
Note: I doubt if it is increasing on (0, + ∞). If so, then M = - 1 is consistent
3. If f (x) is a positive proportional function, then - 5m-3 = 1, i.e. M = - 4 / 5
4. If f (x) is an inverse proportional function, then - 5m-3 = - 1, that is m = - 2 / 5
5. If f (x) is a quadratic function, then - 5m-3 = 2, that is m = - 1
Find the range of the function y = log with the base of (x2-2x + 2)
First look at x ^ 2-2x + 2 = (x-1) ^ 2 + 1 > = 1, so take the logarithm and then Y > = 0, that is, the value range is Y > = 0
∵ constant X & # 178; - 2x + 2 = (x-1) &# 178; + 1 ≥ 1
That is to say, there is always X & # 178; - 2x + 2 ≥ 1
When the base number a > 1,
The range of the function is [0, + ∞)
When the base number 0 < a < 1,
The range of the function is (- ∞, 0]
Given the function f (x) = (m ^ 2-m-1) x ^ (- 5m-3), what is the value of M, f (x): 1. It is a power function, 2. It is a power function and an increasing function on (0, positive infinity), 3. It is a positive proportion function, 4. It is an inverse proportion function, 5. It is a quadratic function,
It seems wrong
When the coefficient m ^ 2-m-1 is not equal to 0, it is a power function, that is, M is not equal to 1 ± √ 2
When the index is - 5m-3 & gt; 0, it increases in the interval, i.e. M & lt; - 3 / 5
When the exponent - 5m-3 = 1, the positive proportion function is m = - 4 / 5
When the exponent - 5m-3 = - 1, the inverse proportion function is m = - 2 / 5
When the exponent - 5m-3 = 2, the quadratic function is m = - 1
m^2-m-1=m^2-m+1/4-1/4-1=(m-1/2)^2-5/4
1: - 5m-3 is not equal to 0 and m ^ 2-m-1 is not equal to 0 = > m is not equal to - 3 / 5 and M is not equal to (1 + radical 5) / 2 and M is not equal to (1-radical 5) / 2
2: (m ^ 2-m-1 > 0 and - 5m-3 > 0) or (m ^ 2-m-1 (1 + radical 5) / 2 and M0) or (m ^ 2-m-1 (1 + radical 5) / 2 and M
It is known that f (x) = log half is the bottom, and the true number is x2 &; MX &; m ① if the value range of function f (x) is r, the value range of real number m is obtained;
∵ f (x) is r, let g (x) = x2-mx-m,
Then G (x) takes all positive numbers
Δ 0 + 4m, i.e. = 4m
Why is he greater than or equal to 0, not less than or equal to 0?
A:
The range of F (x) = Log1 / 2 (x ^ 2-mx-m) is r
The function g (x) = x ^ 2-mx-m has at least one zero point
So: discriminant = (- M) ^ 2-4 * 1 * (- M) > = 0
So: m ^ 2 + 4m > = 0
Solution: m=0
Because the opening of the image of G (x) is upward, if △ 0, then the image of G (x) has no intersection with the X axis, then there must be some positive values g (x) that cannot be obtained, but g (x) requires g (x) to take all positive numbers. If △ 0, G (x) can get all positive values, but the domain of F (x) is not r, and the domain of F (x) is all x values that make g (x) > 0. If G (x) is greater than 0, then G (x) is negative?
The definition field is all x values that let g (x) > 0. Does this affect the range of M? Why? The negative part of G (x) should be rounded off, that is, f (x) should be expanded when G (x) is less than or
Because the opening of the image of G (x) is upward, if △ 0, then the image of G (x) has no intersection with the X axis, then there must be some positive values g (x) that cannot be obtained, but g (x) requires g (x) to take all positive numbers. If △ 0, G (x) can get all positive values, but the domain of F (x) is not r, and the domain of F (x) is all x values that make g (x) > 0. Question: when G (x) is greater than 0, is it negative?
The definition field is all x values that let g (x) > 0. Does this affect the range of M? Why?