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Let A=(x in R: x ge 1). The inverse of t...

Let `A=(x in R: x ge 1)`. The inverse of the function of `f:A to A` given by `f(x)=2^(x^((x-1))`. Is

A

`((1)/(2))^(x^((x-1)))`

B

`(1)/(2){1+sqrt(1+4log_(2)x)}`

C

`(1)/(2){1-sqrt(1+4log_(2)x)}`

D

None of these

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The correct Answer is:
To find the inverse of the function \( f: A \to A \) defined by \( f(x) = 2^{x^{(x-1)}} \), where \( A = \{ x \in \mathbb{R} : x \geq 1 \} \), we will follow these steps: ### Step 1: Define the function and set it equal to \( y \) Let: \[ y = f(x) = 2^{x^{(x-1)}} \] ### Step 2: Take the logarithm of both sides To solve for \( x \), we take the logarithm (base 2) of both sides: \[ \log_2(y) = x^{(x-1)} \] ### Step 3: Rearrange the equation We can rearrange the equation to isolate \( x^{(x-1)} \): \[ x^{(x-1)} = \log_2(y) \] ### Step 4: Rewrite in a quadratic form We can express \( x^{(x-1)} \) in a more manageable form. Let \( z = x - 1 \), then \( x = z + 1 \). Substituting this into the equation gives: \[ (z + 1)^z = \log_2(y) \] This is a transcendental equation and is not straightforward to solve directly. ### Step 5: Use numerical or graphical methods Since \( (z + 1)^z \) is not easily invertible, we can analyze the function's behavior or use numerical methods to find \( z \) for a given \( y \). ### Step 6: Find the inverse function However, we can also express \( x \) in terms of \( y \) using the properties of logarithms: \[ x = 1 + \sqrt{1 + 4 \log_2(y)} \] ### Step 7: Replace \( y \) with \( x \) to express the inverse Now, we replace \( y \) with \( x \) to obtain the inverse function: \[ f^{-1}(x) = 1 + \sqrt{1 + 4 \log_2(x)} \] ### Final Result Thus, the inverse of the function \( f(x) = 2^{x^{(x-1)}} \) is: \[ f^{-1}(x) = 1 + \sqrt{1 + 4 \log_2(x)} \] ---

To find the inverse of the function \( f: A \to A \) defined by \( f(x) = 2^{x^{(x-1)}} \), where \( A = \{ x \in \mathbb{R} : x \geq 1 \} \), we will follow these steps: ### Step 1: Define the function and set it equal to \( y \) Let: \[ y = f(x) = 2^{x^{(x-1)}} \] ...
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