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Let f(x) be a real value function such that f(x)=2x−1x−2∀ x ∈ (2,∞) and g(x)=x2+1x+(f(x))2+1f(x)∀ x>2, then minimum value of g(x) is

Question

Let f(x) be a real value function such that f(x)=2x−1x−2∀ x ∈ (2,∞) and g(x)=x2+1x+(f(x))2+1f(x)∀ x>2, then minimum value of g(x) is
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Solution 1

To find the minimum value of g(x), we first need to simplify the function g(x).

Given that f(x) = 2x - 1/(x - 2), we can substitute f(x) into g(x) to get:

g(x) = x^2 + 1/x + (2x - 1/(x - 2))^2 + 1/(2x - 1/(x - 2))

Next, we need to find the derivative of g(x) to find the critical points. The crit Knowee AI is a powerful AI-powered study tool designed to help you to solve study problem. 

Knowee AI  is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI  is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI  is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI  is a powerful AI-powered study tool designed to help you to solve study problem.
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Let f(x) be a real value function such that f(x)=2x−1x−2∀ x ∈ (2,∞) and g(x)=x2+1x+(f(x))2+1f(x)∀ x>2, then minimum value of g(x) is

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