Find the inverse function of the function y = e ^ x-e ^ - X / 2? I am not smart enough, so I can only trouble the students who know the answer to write the answer in detail. Thank you The standard answer is y = in (x + Geng 1 + x ^ 2)

Find the inverse function of the function y = e ^ x-e ^ - X / 2? I am not smart enough, so I can only trouble the students who know the answer to write the answer in detail. Thank you The standard answer is y = in (x + Geng 1 + x ^ 2)

The solution is as follows.

The inverse function of the function y = 2-x + 1 (x > 0) is () A. y=log21 x−1  （x∈（1，2） B. y=-log21 x−1  （x∈（1，2）） C. y=log21 x−1  （x∈（1，2]） D. y=-log21 x−1  （x（1，2]）

∵ function y = 2-x + 1, x > 0,
∴1＜y＜2．
2-x=y-1，
Take the logarithm base 2 on both sides,
X = log2 (Y-1),
∴x=-log2（y-1），
x. The inverse function of x > 0 is y = - log2 (x-1) = log21
x−1，x∈（1，2）．
Therefore, a

Is the inverse function of the function y = {(0.2) ^ (- x)} + 1? A.y=(log5 x)+1 B.y=(logx 5)+1 C.y=log5 (x-1) D.(log5 x)-1 That's it

Didn't you just ask
y=-log(1/5)(x-1)=log5(x-1)
It's a little bit out of shape? The answer is C

The inverse function of the function y = - x (2 + x) (x ≥ 0)

You can replace x with y, y with X, and then standardize it
y=（1-x)^1/2-1

How to find the inverse function of y = x ^ 2 + X-1 (x > = - 1 / 2)?

y=x^2+x-1
x^2+x=y+1
x^2+x+1/4=y+5/4
(x+1/2)^2=y+5/4
X + 1 / 2 = radical (y + 5 / 4)
X = radical (y + 5 / 4) - 1 / 2
Because x > = - 1 / 2
So Y > = - 5 / 4
So, the inverse of the function is
Y = radical (x + 5 / 2) - 1 / 2 (x > = - 5 / 4)

If the inverse function F-1 (x) = 1 + X (x < 0), then f (2)=

If the inverse function F-1 (x) = 1 + x ^ 2, then f (2) is expressed as 1 + x ^ 2 = 2 x ^ 2 = 1 and X

The solution equation is: (2x + 3) (x-4) - (x + 2) (x-3) = x2 + 6

（2x+3）（x-4）-（x+2）（x-3）=x2+6，
By simplification, we get (2x2-5x-12) - (x2-x-6) = x2 + 6,
Remove the brackets and get 2x2-5x-12-x2 + X + 6 = x2 + 6,
By moving the term, - 4x = 12,
X = - 3

(2x-1) (2x + 1) - 3 (x + 2) (X-2) = (x + 1) (X-2) + 12 mathematical problems,

4x^2-1-3x^2+12=x^2-x+10
x^2+11=x^2-x+10
10-x=11
x=-1

solve equations: （1）5 2x+3=3 x−1； （2）x x−1＝3 2x−2−2．

0

0

1/4(2x+3)^3=2
2X + 3 is not in the denominator, right
（2x+3）³=8
2x+3=2
2x= -1
x= -1/2