If f(x) = x left( frac{1}{x-1} + frac{1}{x} + | vedclass

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(D) Given function is $f(x) = x \left( \frac{1}{x-1} + \frac{1}{x} + \frac{1}{x+1} \right)$ for $x > 1$. Simplifying the expression inside the bracket

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$f(x) > 3$

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(D) Given function is $f(x) = x \left( \frac{1}{x-1} + \frac{1}{x} + \frac{1}{x+1} \right)$ for $x > 1$. Simplifying the expression inside the bracket: $f(x) = x \left( \frac{(x+1) + (x-1)}{(x-1)(x+1)} + \frac{1}{x} \right)$ $f(x) = x \left( \frac{2x}{x^2 - 1} + \frac{1}{x} \right)$ $f(x) = \frac{2x^2}{x^2 - 1} + 1$ $f(x) = \frac{2}{1 - \frac{1}{x^2}} + 1$ Since $x > 1$, we have $x^2 > 1$, which implies $0 Therefore, $0  1$. Multiplying by $2$, we get $\frac{2}{1 - \frac{1}{x^2}} > 2$. Adding $1$ to both sides, we get $f(x) > 2 + 1 = 3$. Thus, $f(x) > 3$.

If $f(x) = x \left( \frac{1}{x-1} + \frac{1}{x} + \frac{1}{x+1} \right)$ for $x > 1$, then:

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