The sequence {an} is an arithmetic sequence and Sn is the sum of the first n terms If S7 = 8, S8 = 7, then S15=______

The sequence {an} is an arithmetic sequence and Sn is the sum of the first n terms If S7 = 8, S8 = 7, then S15=______

a8=s8-s7=-1
s15=a8*15=-15

The teacher said It's not right to directly convert f (x + 1) to f (- 1) (x + 1). How to make it direct It should be f (x + 1) - - f (x) - - f (- 1) (x) - - f (- 1) (x + 1) Can you give me a specific example~

If f (x + 1) = x ^ 3 + 1, the inverse function can't be the third root x plus 1 directly
It should be changed first so that x + 1 = t, that is, x + 1 is regarded as a whole, because f (x + 1) is a function of X + 1, not about X
The discussion of inverse function is about the whole thing inside the bracket,

Finding the inverse function of F (x-1) with known f (x) Should the idea be f (x) - f (x-1) - f '(x-1) or F (x) - f' (x) - f '(x-1)? Finding the inverse function of F (x + 1) with known f (x + 1) Does it need to be changed to f (x)?

Take an example from junior high school, such as finding the inverse function of F (x) = y = 2x
X = Y / 2. The inverse function is required. Just replace X and y
So f '(x) = x / 2

Evaluation: cos (2arctan2)=_________

Let x = arctan2
tanx=2
tan2x=2tanx/(1-tan^2x)=-4/3
Because x ∈ (0, Π / 2)
2x∈(0,∏)
So sin2x > 0
Because tan2x = sin2x / cos2x

How to calculate the domain of Ln (- 4 + 2ln (1 + 2 * x)) and its inverse function

-4 + 2ln (1 + 2 * x) > 0, 1 + 2 * x > 0

Find the definition domain of this inverse function y = arccos2x / (1 + x)

Is the definition domain of inverse trigonometric function y = arccos [2x / (1 + x)]?
-1 ≤ 2x / (1 + x) ≤ 1, that is: - 1 ≤ [2 (x + 1) - 2] / (x + 1) ≤ 1, that is: - 1 ≤ 2-2 / (x + 1) ≤ 1,
In other words: - 3 ≤ - 2 / (x + 1) ≤ - 1, that is: 1 / 2 ≤ 1 / (x + 1) ≤ 3 / 2; therefore, 2 / 3 ≤ x + 1 ≤ 2;
The result is: - 1 / 3 ≤ x ≤ 1
Therefore, the definition domain is: [- 1 / 3,1]
If you don't understand, please hi me,

The definition domain and monotone interval of the following functions are pointed out, and the inverse function on the single point interval is obtained （1）f（x）=√（2x-1）；（2）f（x）=-1/x+1；（3）f（x）=x²+8

（1）f（x）=√（2x-1）；
Therefore, it is defined as the interval [2 +] / 2 = 1 / 2;
Therefore, the inverse function is: F (x) = (y 2 + 1) / 2;
（2）f（x）=-1/x+1；
Here, X is the denominator of which is not zero. The definition domain is x ≠ 0, monotone interval (- ∞, 0) ∪ (0, + ∞); y = - 1 / x + 1, x = - 1 / (Y-1), and the inverse function is: F (x) = - 1 / (x-1);
（3）f（x）=x²+8
The definition domain is x ≠ 0, monotone interval (- ∞, 0) ∪ (0, + ∞); y = x ^ + 8, x ^ = Y-8, x = √ (Y-8), and the inverse function is: F (x) = √ (X-8)

Whether the monotone function on the definition field must have an inverse function, and whether the non monotonic function on the definition field must have no inverse function, please give an example Please give specific examples,

If monotone is defined on a domain, a single valued function must have an inverse function, and its inverse function must also be monotone
f(x)=x
A nonmonotone function on a domain of definition is not necessarily without an inverse function
But it's not an elementary function
Can be a segmented constructor

The function f (x) = (AX + 1) / (4x + 3) in senior one mathematics has inverse function in the domain of definition. Find the value range of A hurry!

A is not equal to plus or minus four thirds
Find the inverse function so that it is not equal to minus three quarters
In the original equation, according to the significance of the inverse function, it is concluded that f (x) is not equal to a quarter of A
Solve two equations

y=4cos(x/3),0

y=4cos(x/3)
y/4=cos(x/3)
x/3=arccos(y/4)
x=3arccos(y/4)
Exchange variables
The inverse function is y = 3 arccos (x / 4)
Because the original function y = 4cos (x / 3), 0