If f (x-1) = x & # 178; - 2x-3, then f (x) If f (x-1) = x & # 178; - 2x-3, then f (x)=

If f (x-1) = x & # 178; - 2x-3, then f (x) If f (x-1) = x & # 178; - 2x-3, then f (x)=

f(x-1)=x^2-2x-3
Let X-1 = t
x=t+1
Then f (T) = (T + 1) ^ 2-2 (T + 1) - 3
=(t+1)(t+1-2)-3
=(t+1)(t-1)-3
=t^2-1-3
=t^2-4
Let t = X
Then f (x) = x ^ 2-4

If f (x-1) = x & # 178; - 2x + 3 (x ≤ 1), then f ^ - 1 (4) is equal to

Let f (x-1) = x & # 178; - 2x + 3 = 4
x

Fraction! X & # 178; - 2x + 1 / X & # 178; - x

Factorization?
The original formula = (x-1) & # / X (x-1) = X-1 / X
Just check the sum of squares and extract the common factor
Hope to help you

How to calculate the domain of X if the square of x plus 1 is not equal to 0

∵ X & # 178; ≥ 0 ∵ X & # 178; + 1 > 0, i.e. X & # 178; + 1 is not equal to 0; when x is any value, X & # 178; + 1 is not equal to 0

Calculation of machine number by computer
If a binary integer with a length of 1 byte is represented by a complement and consists of 5 1s and 3 0s, the minimum decimal integer that can be represented is?

Complement: 10001111
Original code: 11110001 = - 113
The first bit of the complement is the sign bit. Since the complement is obtained by adding 1 to the original code, the other four 1s are placed in the last four bits. Because after changing back to the original code, the higher the 1 is, the smaller the number is
I don't know, right? For your reference

It is known that a = 1 + 1 / 2 + 1 / 3 + +1/2012,B=1/2+1/3+… +1 / 2013. Seeking a-b

A-B=1-1/2013=2012/2013

For the differentiable function f (x) defined on R, if the image of y = EF '(x) is shown in the figure, then the increasing interval of y = f (x) is______ ．

The interval f '(x) ≥ 0 is (- ∞, 2), so the increasing interval of function y = f (x) (- ∞, 2), so the answer is: (- ∞, 2),

Calculation: (1 - √ 2) + (√ 2 - √ 3) + (√ 3 - √ 4) + +（√2004-√2005）

Get rid of the brackets
Original formula = 1 - √ 2 + √ 2 - √ 3 + √ 3 - √ 4 + +√2004-√2005
=1-√2005

After translating point P (2, - 1) to the right for 3 unit lengths, the coordinates of point P1 are obtained______ ．

∵ point P (2, - 1) shifts 3 unit lengths to the right and arrives at point P1. The abscissa of P1 is 2 + 3 = 5 and the ordinate is - 1, so the answer is: (5, - 1)

X = one third by four ninths = one sixth (solving the equation)

1 has 24 points to 3