Why to find the increasing interval of logarithmic function is to find the decreasing interval of compound function

Why to find the increasing interval of logarithmic function is to find the decreasing interval of compound function

For a logarithmic function, for example, if it is ln (- x), its domain of definition should be the part that makes (- x) > 0, that is, the part of x0, that is, x > - 1. Therefore, it is not necessary to reduce the interval, but to solve it according to the function of independent variable

1. Is the image of y = log (1 / 3) x ^ 2 the same as that of y = 2log (1 / 3) x? (the first bracket is the base)
2. Given that f (x) = log (a) (x + 1) is [0,1], then the value of a is equal to?
A. 2 b. radical 2 C. half radical 2 D one third

1) The definition field of y = log (1 / 3) x ^ 2 is a real number except 0, and the definition field of y = 2log (1 / 3) x is x > 0, so the image is different. 2) the definition field and value field of function f (x) = log (a) (x + 1) are [0,1], that is, 1 ≤ x + 1 ≤ 2, and the value field of F (x) = log (a) (x + 1) is [0,1]. Obviously, a > 1, if a = radical 2, then the value field is

Loga [(T + 1) / 2) - 1 / 2 loga (T) how to get the root of loga [(T + 1) / 2 T] should be understood
I don't understand. Is there a higher level of help

1 / 2 loga (T) = loga [(T + 1) / 2's root sign T, 1 / 2's exponent from logarithm coefficient to true number, which is the basic transformation
The subtraction of two logarithms of the same base is equal to the division of their true numbers by the root of (T + 1) / 2

The number of zeros of function f (x) = lgx + 2x-7 is () and the interval is ()

There is only one zero point. It is an increasing function. Its approximate interval is (3,4)

The sum of all zeros of function f (x) = {x ^ 2 + 2x-3, X ≤ 0, lgx-1, x > 0 is
x^2+2x-3x≤0,
F (x) = {lgx-1, x > 0, the sum of all zeros is

x≤0
x²+2x-3=0
(x+3)(x-1)=0
So x = - 3
x>0
lgx-1=0
lgx=1
x=10
So sum = - 3 + 10 = 7

Given that f (x) = x + m / X (M belongs to R), if M = 2, find the maximum value of the function g (x) = f (x) - INX in the interval [1,3 / 2]

The maximum value of F (x) = x + 2 / X in the interval [1,3 / 2] is the largest of F (1) and f (3 / 2)
f(1)=3 f(3/2)=3/2+4/3=17/6
Therefore, the maximum value of F (x) is f (1) = 3

If the maximum value of function y = - (sin α - M) &# 178; + M & # 178; - 2m - 1 (0 less than or equal to α, less than or equal to π / 2) is negative, the value range of M is obtained

0 ≤ α ≤ π / 2, then 0 ≤ sin α ≤ 1, when m < 0, sin α = 0, y takes the maximum value - (0-m) &# 178; + M & # 178; - 2m-1 = - 2m-1 ＜ 0, then m ＞ - 1 / 2, combined with the premise m < 0, that is - 1 / 2 ＜ m ＜ 0, which satisfies the meaning; when 0 ≤ m ≤ 1, sin α = m, y takes the maximum value M & # 178; - 2m-1 = (m-1) ^ 2-2 ＜

If M ＞ 2, then f (θ) = sin2 θ + MCOs θ, θ∈ the maximum value of R G (m)=______ ．

From the knowledge of trigonometric function, we can get f (θ) = sin2 θ + MCOs θ = - Cos2 θ + MCOs θ + 1, let cos θ = t, then t ∈ [- 1, 1] can be transformed into y = - T2 + MT + 1, t ∈ [- 1, 1] formula can get y = - (t − m2) 2 + 1 + M24, we can know that the function image of T is a parabola with opening downward, symmetry axis is t = m2, and M > 2, so M2 > 1, so the function increases monotonically in [- 1, 1], so g (m) = - 12 + m × 1 + 1 = M So the answer is: M

If M ＞ 2, then f (θ) = sin2 θ + MCOs θ, θ∈ the maximum value of R G (m)=______ ．

From the knowledge of trigonometric function, we can get f (θ) = sin2 θ + MCOs θ = - Cos2 θ + MCOs θ + 1, let cos θ = t, then t ∈ [- 1, 1] can be transformed into y = - T2 + MT + 1, t ∈ [- 1, 1] formula can get y = - (t − m2) 2 + 1 + M24, we can know that the function image of T is a parabola with opening downward, symmetry axis is t = m2, and M > 2, so M2 > 1, so the function increases monotonically in [- 1, 1], so g (m) = - 12 + m × 1 + 1 = M So the answer is: M

F (x) is an odd function. When x > 0, f (x) = log2 ^ x, find X

x0
therefore
f(-x)=log2^(-x)
So - f (x) = log2 ^ (- x)
So when x