Finding monotone interval of function, f = log3 X-1 / 2x + 1 0.0.

Finding monotone interval of function, f = log3 X-1 / 2x + 1 0.0.

The true number is greater than 0
(2x+1)/(x-1)>0
That is, (2x + 1) (x-1) > 0
X1
So it's (- ∞, - 1 / 2) ∪ (1, + ∞)
The range of the function f (x) = - 2 to the power of 2x + 1 + 5 × 2 to the power of X + 1, X ∈ (- infinity, 0) is
F (x) = - 2 ^ (2x + 1) + 5 * 2 ^ x + 1 = - 2 * (2 ^ x) ^ 2 + 5 * 2 ^ x + 1, let t = 2 ^ x, because x ∈ (- ∞, 0), so t belongs to (0,1), so f (x) and y = - 2 (T-5 / 4) ^ 2 + 1 + 25 / 8 = - 2 (T-5 / 4) ^ 2 + 33 / 8 have the same range, y = - 2 (T-5 / 4) ^ 2 + 33 / 8 is the parabola with the opening downward, 1 < y < 4, that is, the range of F (x) is (1,4)
The monotone increasing interval of function f (x) = log3 (x2-2x-8) is
（4,+∞）
Given that 0 is less than X and less than or equal to 1 / 4, find the range of function y = x ^ 2-2x + 2 / X
y=(x^2-2x+2)/x
Is it y = (x ^ 2-2x + 2) / X or y = (x ^ 2-2x) + 2 / x? Write it clearly. The two results are different
Monotone increasing interval of function y = √ 4-3x-x ^ 2
Such as the title
√（4-3x-x^2）
Domain of definition
4-3x-x^2>=0
D-4
Given that 2 ^ (2x) is less than or equal to (1 / 4) ^ x, find the range of function y = 2 ^ X
Given that 2 ^ (2x) is less than or equal to (1 / 4) ^ x, find the range of function y = 2 ^ X
2^2x
From the condition: X
What is the monotone increasing interval of exponent-x ^ 2 + 3x-2 of function y = 2?
Y = a ^ x, when a > 1, x > 0, y increases with the increase of X
When 00, - x ^ 2 + 3x-2 is an increasing function, so the interval of - 2x + 3 > 0 is [- ∞, 3 / 2]. However, here x is more than 1
So: so the intersection is (1,3 / 2]
Since the x power of y = 2 is an increasing function, we only need to find the monotone increasing interval of - x ^ 2 + 3x-2, and at the same time, we should ensure that - x ^ 2 + 3x-2 > 0. -The monotone increasing interval of x ^ 2 + 3x-2 is that the solution of x0 is 1
Y = 2x (1 / 2 less than or equal to x less than or equal to 4) to find the range of function
Y = 2x increasing
So x = 1 / 2, minimum y = 1
X = 4, maximum y = 8
Range [1,8]
1/2
Monotone increasing interval of function y = | (x ^ 2) - 3x + 2 | D
Draw the image of y = (x ^ 2) - 3x + 2, and then fold the symmetry below the Y axis over the Y axis
It is easy to see that the simple increase is [1,1.5] and [2, positive infinity]
[1,1.5] and {x | x > 2}
Let f (x) = x & # 178; - 2x + 1-k & # 178;; (k belongs to R), if f (x) > 2k-2 is constant for any x belonging to (0, + infinity), the value range of K is obtained
f(x)=x²-2x+1-k²＞2k-2;
x²-2x+1-k²-2k-1+3＞0;
(x-1) &# 178; - (K + 2) &# 178; + 3 > 0
∴k=-2；
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Solution;
f(x)=x²-2x+1-k²=(x-1)²-k²
f(x)>2k-2
It holds for any x ∈ (0, + infinity)
Namely
(x-1)²>k²+2k-2
It holds for any x ∈ (0, + infinity)
The minimum value of (x-1) &# 178 in (0, + infinity) is: 0
∴0>k²+2k-2
That is K & # 178; + 2k-2 (K + 1) &# 178; - 3
-√3