The real valued function f(x) = frac{x}{e^x - | vedclass

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(B) Given $f(x) = \frac{x}{e^x - 1} + \frac{x}{2} + 1$. To check if the function is even or odd,we evaluate $f(-x)$. $f(-x) = \frac{-x}{e^{-x} - 1} +

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An even function

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(B) Given $f(x) = \frac{x}{e^x - 1} + \frac{x}{2} + 1$. To check if the function is even or odd,we evaluate $f(-x)$. $f(-x) = \frac{-x}{e^{-x} - 1} + \frac{-x}{2} + 1$. Multiply the numerator and denominator of the first term by $e^x$: $f(-x) = \frac{-x e^x}{1 - e^x} - \frac{x}{2} + 1 = \frac{x e^x}{e^x - 1} - \frac{x}{2} + 1$. We can rewrite $\frac{x e^x}{e^x - 1}$ as $\frac{x(e^x - 1 + 1)}{e^x - 1} = x + \frac{x}{e^x - 1}$. Substituting this back: $f(-x) = x + \frac{x}{e^x - 1} - \frac{x}{2} + 1 = \frac{x}{e^x - 1} + \frac{x}{2} + 1$. Since $f(-x) = f(x)$,the function is an even function.

The real valued function $f(x) = \frac{x}{e^x - 1} + \frac{x}{2} + 1$ defined on $R \backslash \{0\}$ is

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