The known function f (x2 − 3) = lgx2 x2−6． (1) The definition domain of function f (x) is obtained; (2) the parity of function f (x) is judged; (3) Find the inverse function of F (x); (4) if f [φ (x)] = lgx, find the value of φ (3)

(1) Let x2-3 = t (t ＞ - 3), so the original function is transformed into f (T) = LG T + 3T − 3. From t + 3T − 3 > 0, the definition domain is {t| T > 3}, that is, f (x) = LG x + 3x − 3, and the definition domain is {x | x ＞ 3} (2) because the definition domain of F (x) is (3, + ∞), so the function f (x) is non odd and non even function (3)

y=4cosx/3,0

y=4cosx/3
x/3=arccos(y/4)
x=3arccos(y/4)
The inverse function is: y = 3 arccos (x / 4)

Find the definition domain and value range of the following functions: （1）y=log2（x-2）；（2）y=log4（x2+8）．

(1) From X-2 > 0, x > 2,
Therefore, the definition domain of the function y = log2 (X-2) is (2, + ∞), and the range is r
(2) Because for any real number x, log4 (x2 + 8) makes sense,
So the definition domain of the function y = log4 (x2 + 8) is r
Because x2 + 8 ≥ 8,
So log4 (x2 + 8) ≥ log48 = 3
2，
That is, the value range of the function y = log4 (x2 + 8) is [3]
2，+∞）．

Find the definition domain of inverse function of F (x) = log2 (x + 1) (x > = 4) As the title!

The definition domain of inverse function is the range of original function
F (x) monotonically increasing
therefore
Definition domain of inverse function [log2 ^ 5, + ∞)

Given the domain of F (x) definition, how to find the range without inverse function (the more detailed the better) It's OK to have a classified discussion

I think the inverse function is the most stupid way to solve the range problem
First of all, f (x) is corresponding to a formula
If it is a quadratic function, it can be formulated, or the result can be obtained directly from the image
Of course, quadratic function is relatively simple. For some special functions, such as trigonometric function, we need to consider the definition of many points. First change to form a general form, draw the image, combined with the definition domain can also get the result
If it is difficult to draw some images, such as y = ax + B / X
We can use the mean value theorem to find the maximum (minimum) value [mean value theorem: a + B ≥ 2 √ AB, (a > 0, b > 0)]. Then look at the conditions for obtaining the maximum value. If it is the above formula, that is, ax = B / x, this is the common method when the definition domain is r, and you can directly see the value range
If it is a composite function or a piecewise function, we should start with monotone interval. We can use derivative to find monotone interval. If we don't have derivative, we will discuss it
Composite function, the same increase but different decrease, that is, if the monotonicity of two composite functions is the same, the original function is increasing, if the monotonicity is different, the original function is decreasing
For example, f (x) = (SiNx) ^ 2 (domain: [0, д / 2])
This is a compound function of trigonometric function and quadratic function
In the definition domain, SiNx is increasing, and when SiNx ≥ 0 (SiNx) ^ 2, it is also increasing. And in the definition domain, the monotonicity of the two is the same, so the original function is an increasing function. Knowing the monotonicity and the definition domain, it is good to find the value range
For example, quadratic function, logarithmic function, trigonometric function

The definition domain of inverse function is obtained from the original function range, For example, the inverse function of F (x) = LG (2-x) / (2 + x) is y = 2 (1-10 ^ x) / (1 + 10 ^ x), so how to find its definition domain is very difficult

The definition domain is the range of independent variables that can make a function meaningful. The definition domain of the function you give is r

Is the natural domain of the inverse function of a function the value range of the original function? Why

A function has an inverse function. First of all, it is a monotone function. The original function has a unique function value corresponding to any independent variable. In essence, an inverse function has a unique independent variable corresponding to any function value of the original function. It uses y to represent x, and finally adjusts the position of X and y

Seeking the range of value and definition of inverse function in two questions The definition field of condition cos (x) is [0, Pai]. Find the definition field and value field of COS inverse (x ^ 2) I found that the definition field is [- 1,1], but why is the value field not [0, Pai], but [0, Pai / 2]? Condition: the domain of Tan (x) is (- Pai / 2, Pai / 2). Find the domain and value field of Tan inverse (2x Pai / 2) This domain is r, I know, but why is the answer to the range (0, Pai / 2), not (- pi / 2, Pai / 2)? Definition domain of COS (x ^ 2): 0 "x ^ 2" π ν x ∈ [- radical sign π, root sign π] This statement contradicts the definition of line and plane from - 1 to 1. And the range of arccosx can only be [0, π], is it absolute? Is it true that no matter what the range of the original function is, it will always be zero to pie?

∵ the definition domain of inverse function is the range of original function, and the range of inverse function is the domain of original function
The definition domain of COS (x ^ 2): 0 "x ^ 2" π ν x ∈ [- root sign π, root sign π]
The value range of COS (x ^ 2) [- 1,1]
But arccosx can only be in the range of [0, π]
Definition domain of COS (x ^ 2 inverse function [- 1,1]
Range [0, radical π]

Given that the image of function y = FX and the image of function y = (x-1) / (x + 1) are symmetric with respect to the line y = x, the analytic formula of function y = FX is obtained

The inverse function of y = (x-1) / (x + 1) is calculated
xy+y=x-1
xy-x=-1-y
x(y-1)=-(1+y)
x=-(1+y)/(y-1)
So y = - (1 + x) / (x-1)

Inverse function of senior one 1. Y = 1-2 / x + 3 {x belongs to the inverse function of R and X is not equal to - 3} 2. Y = 4x + 1 / 5x-3 {x belongs to the inverse function of R and X is not equal to 3 / 5} 3. The inverse function of y = root 2X-4 {x > equal to 2}

① The inverse function of the original function is y = 2 / (1-x) - 3 (x ≠ 1) ② y = 4x + 1 / 5x-3 (5x-3) × y = 4x + 1 (5y-4) x = 3Y + 1 x = 3Y + 1 / (5y-4) / / the inverse function of the original function is y = 3x + 1 / (5x-4) (x ≠ 4 / 5) ③ y = (2X-4

The known function f (x2 − 3) = lgx2 x2−6． (1) The definition domain of function f (x) is obtained; (2) the parity of function f (x) is judged; (3) Find the inverse function of F (x); (4) if f [φ (x)] = lgx, find the value of φ (3)