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
Solution
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 critical points are where the derivative is equal to zero or undefined.
The derivative of g(x) is:
g'(x) = 2x - 1/x^2 + 2*(2x - 1/(x - 2)) - 1/(2x - 1/(x - 2))^2
Setting g'(x) equal to zero and solving for x will give us the critical points.
Once we have the critical points, we can substitute them back into g(x) to find the corresponding y-values. The minimum value of g(x) will be the smallest y-value.
Please note that this is a complex problem that requires knowledge of calculus, specifically differentiation. If you're not familiar with these concepts, you may need to study them further to fully understand the solution.
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