If $f(x)=2\left(\tan ^{-1}\left(e^{x}\right)-\frac{\pi}{4}\right)$, then $f(x)$ is :

We're given:

$f(x) = 2\left(\tan^{-1}(e^x) - \dfrac{\pi}{4}\right)$

Note that $\tan^{-1}(1) = \dfrac{\pi}{4}$, and $e^0 = 1$, so:

$f(0) = 2\left(\tan^{-1}(1) - \dfrac{\pi}{4}\right) = 2\left(\dfrac{\pi}{4} - \dfrac{\pi}{4}\right) = 0$

Since $f(0) = 0$, this hints that $f(x)$ could be an odd function. Let's verify.

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A function is odd if $f(-x) = -f(x)$. Let's compute $f(-x)$:

$f(-x) = 2\left(\tan^{-1}(e^{-x}) - \dfrac{\pi}{4}\right)$

We use the identity: for any $t > 0$,

$\tan^{-1}(t) + \tan^{-1}\left(\dfrac{1}{t}\right) = \dfrac{\pi}{2}$

This works because if $\tan(\alpha) = t$, then $\tan\left(\dfrac{\pi}{2} - \alpha\right) = \dfrac{1}{t}$, so the two angles must add to $\dfrac{\pi}{2}$.

Since $e^x > 0$ for all $x$, we substitute $t = e^x$:

$\tan^{-1}(e^{-x}) = \dfrac{\pi}{2} - \tan^{-1}(e^x)$

Substituting back:

$f(-x) = 2\left(\dfrac{\pi}{2} - \tan^{-1}(e^x) - \dfrac{\pi}{4}\right)$

$= 2\left(\dfrac{\pi}{4} - \tan^{-1}(e^x)\right)$

$= -2\left(\tan^{-1}(e^x) - \dfrac{\pi}{4}\right)$

$= -f(x)$

$\therefore f(x)$ is an odd function.

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Differentiating to check if $f(x)$ is increasing or decreasing:

$f'(x) = 2 \cdot \dfrac{1}{1 + (e^x)^2} \cdot e^x = \dfrac{2e^x}{1 + e^{2x}}$

Since $e^x > 0$ and $1 + e^{2x} > 0$ for all $x$, we have $f'(x) > 0$ for all $x$.

$\therefore f(x)$ is strictly increasing.

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As $x \to \infty$: $e^x \to \infty$, so $\tan^{-1}(e^x) \to \dfrac{\pi}{2}$, giving:

$f(x) \to 2\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}\right) = \dfrac{\pi}{2}$

As $x \to -\infty$: $e^x \to 0$, so $\tan^{-1}(e^x) \to 0$, giving:

$f(x) \to 2\left(0 - \dfrac{\pi}{4}\right) = -\dfrac{\pi}{2}$

These values are approached but never reached, so the range is $\left(-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right)$.

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$f(x)$ is an odd, strictly increasing function with range $\left(-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right)$.