High inverse function, There is no inverse function for the power function y = x ^ 2. Why? Detailed explanation, thank you If the power function y = x ^ a (a > 0) has an inverse function, then a can take____ If the power function y = x ^ a (a)

There is no inverse function for the power function y = x ^ 2. Why?
Therefore, an independent variable corresponds to two functions, which does not conform to the definition of function
If the power function y = x ^ a (a > 0) has an inverse function, then a can take_ 3___
If the power function y = x ^ a (a)

Let a > 0 and a not equal to 1. F (x) = log (a, [x + sqrt (x ^ 2-1)]), X ≥ 1, find the inverse function of function f (x), and find the definition domain of this inverse function Note: log (a, b) is the logarithm with a as the base and B as the true number. Sqrt is the square root and x ^ 2 is the square of X The key steps should not be missed

y=A^2X/2A^X-1
X is not equal to 5 / 4

It is known that f (x) = ((x-1) / (x + 1)) ^ 2 (1) Find the inverse function F-1 (x) of F (x) (2) If the inequality (1 - √ x) [F-1 (x)] > A (a - √ x) holds for all x ∈ [1 / 4,1 / 2], find the value range of A I really want guidance

（1）y=((x-1)/(x+1))²
√ y = (x-1) / (x + 1)
√y（x+1）=x-1
x（√y-1）=-√y-1
x=（-√y-1）/（（√y-1）
Therefore, F-1 (x) = - (√ x + 1) / (√ x-1) x ≥ 0 and X ≠ 1
(2) Substituting F-1 (x) into (1 - √ x) ((- √ x-1) / (√ x-1)) > A (a - √ x)
√x+1＞a（a-√x）
The results are as follows: (a + 1) √ x-a 2 + 1 > 0
(consider it as a straight line with √ x as the independent variable, which is greater than 0 at 1 / 4 and 1 / 2)
① By substituting x = 1 / 4, we can get the following result: 2A 2 - A-3 ≤ 0, and the solution: - 1 < a < 3 / 2
② By substituting x = 1 / 2, we can get the following results: 2A 2 - √ 2A - (√ 2 + 1) ≤ 0, the solution is: √ 2 / 2 < a < 1 + √ 2 / 2
Take the intersection to get: √ 2 / 2 < a < 3 / 2
If you don't understand, ask again,

It is known that the function y = f (x) is a function of degree, and f (1) = 1, f [f (2)] = 2F ^ - 1 (4). Find the expression of F (x)! The more wordy the better!... f ^ - 1 is an inverse function, The answer in the book is f (x) = 2x-1!

Let f (x) = KX + B
There is f (1) = K + B = 1 so B = 1-k
f(x)=kx+1-k=k(x-1)+1
so f(2)=k+1
f[f(2)]=f(k+1)=k^2+1=2f^-1(4).
so f^-1(4).=(k^2+1)/2
so f[k^2+1/2)]=k[(k^2-1)/2]+1=4
so k^3-k=6
so k=2
so b=1-2=-1
so f(x)=kx+b=2x-1
Is that enough detail
You can ask me if you don't understand~

Ultrasimple inverse function Find the inverse function of y = (10 ^ x + 10 ^ - x) / (10 ^ X-10 ^ - x) + 1, the specific steps, thank you

From the meaning of the title
y-1=[exp(xln10)+exp(-xln10)]/[exp(xln10)-exp(-xln10)]
=CH (xln10) / sh (xln10), then Y-1 = 1 / {th (xln10)}
Then th (xln10) = 1 / (Y-1) gets xln10 = arth (1 / (Y-1)) and then
X = arth [1 / (Y-1)] / ln10, then the inverse function is
y=arth[1/(x-1)]/ln10

Find a simple inverse function Y=lg(X-1)/(X+1)

Definition domain (x-1) / (x + 1) > 0
So x > 1 or X

If the function y = f (x) is the inverse function of the function y = ax (a > 0 and a ≠ 1), its image passes through a point（ a. A), then f (x) =___ .

The inverse function of ∵ function y = ax is f (x) = logax（
a，a），
∴a=loga
a. That is, a = 1
2，
So the answer is: Log1
2x．

Is the inverse function of the function y = radical x + 1? Quick return

y = √(x + 1)
Definition domain x ≥ - 1
Range y ≥ 0
y = √(x + 1)
y² = x + 1
x = y² - 1
So the inverse function is y = x? - 1 (x ≥ 0)

Note that the inverse function of the function y = f (x) is the negative first power (x) of y = F. if the image of function y = f (x) crosses the point (1,0), how many points does the image of its inverse function pass through

（0,1）

The inverse function of the function y = e to the power of X + 1?

Y = e ^ x + 1, the range is Y > 1, the inverse solution, e ^ x = Y-1,
The natural logarithm is taken on both sides,
X = ln (Y-1), x = ln (Y-1), x = ln (x-1) is obtained by exchanging X and y, and the definition domain is x > 1

High inverse function, There is no inverse function for the power function y = x ^ 2. Why? Detailed explanation, thank you If the power function y = x ^ a (a > 0) has an inverse function, then a can take____ If the power function y = x ^ a (a)