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Given the function $f(x) = e^x - e^{-x} + x - \tan^{-1}x$, we need to find its derivative $f'(x)$.
Differentiating each term with respect to $x$, we get: $f'(x) = \frac{d}{dx}(e^x) - \frac{d}{dx}(e^{-x}) + \frac{d}{dx}(x) - \frac{d}{dx}(\tan^{-1}x)$ $f'(x) = e^x - (-e^{-x}) + 1 - \frac{1}{1+x^2}$ $f'(x) = e^x + e^{-x} + 1 - \frac{1}{1+x^2}$
We can rewrite $e^x + e^{-x}$ as $2\cosh(x)$, but it's not necessary for this problem. Let's analyze the expression $f'(x) = e^x + e^{-x} + 1 - \frac{1}{1+x^2}$.
We know that $e^x > 0$ and $e^{-x} > 0$ for all $x$. Also, $e^x + e^{-x} \geq 2$ (since the minimum value of $e^x + e^{-x}$ occurs at $x=0$, where $e^0 + e^0 = 1+1 = 2$). Now, consider the term $1 - \frac{1}{1+x^2}$. Since $x^2 \geq 0$, we have $1+x^2 \geq 1$, so $0 < \frac{1}{1+x^2} \leq 1$. Therefore, $1 - \frac{1}{1+x^2} \geq 0$. Thus, $f'(x) = e^x + e^{-x} + 1 - \frac{1}{1+x^2} \geq 2 + 0 = 2 > 0$. Since $f'(x) > 0$ for all $x$ in the domain of $f(x)$, the function $f(x)$ is strictly increasing.
Final Answer: f(x) is strictly increasing.
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