When analyzing the parity of a function, check if \( f(-x) = f(x) \) (even) or \( f(-x) = -f(x) \) (odd). For monotonicity, compute the derivative of the function: if \( f'(x) > 0 \), the function is strictly increasing, and if \( f'(x) < 0 \), the function is strictly decreasing. A positive derivative over the entire domain typically indicates an increasing function.
To classify the function \( f(x) = 2\left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \), we examine its symmetry (even/odd) and its rate of change (monotonicity).
1. Symmetry Analysis:
A function is odd if \( f(-x) = -f(x) \) for all \( x \). Evaluating \( f(-x) \):
\( f(-x) = 2\left( \tan^{-1}(e^{-x}) - \frac{\pi}{4} \right) \)
Using the identity \( \tan^{-1}(e^{-x}) = \frac{\pi}{2} - \tan^{-1}(e^x) \), we get:
\( f(-x) = 2\left( \left( \frac{\pi}{2} - \tan^{-1}(e^x) \right) - \frac{\pi}{4} \right) = 2\left( \frac{\pi}{4} - \tan^{-1}(e^x) \right)\)
\( f(-x) = \frac{\pi}{2} - 2\tan^{-1}(e^x) = -\left( 2\tan^{-1}(e^x) - \frac{\pi}{2} \right)\)
This does not directly match \(-f(x)\). Let's recheck the calculation of \( f(-x) \).
\( f(-x) = 2\left( \left( \frac{\pi}{2} - \tan^{-1}(e^x) \right) - \frac{\pi}{4} \right) = 2\left( \frac{\pi}{2} - \frac{\pi}{4} - \tan^{-1}(e^x) \right) = 2\left( \frac{\pi}{4} - \tan^{-1}(e^x) \right)\)
This simplifies to \( f(-x) = \frac{\pi}{2} - 2\tan^{-1}(e^x) \). Now let's compare with \(-f(x)\):
\(-f(x) = -2\left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) = -2\tan^{-1}(e^x) + \frac{\pi}{2}\)
Therefore, \( f(-x) = -f(x) \), confirming that \( f(x) \) is an odd function.
2. Monotonicity Analysis:
To determine if the function is increasing or decreasing, we compute its first derivative, \( f'(x) \).
\( f(x) = 2\tan^{-1}(e^x) - \frac{\pi}{2} \)
The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1+u^2} \cdot \frac{du}{dx} \). For \( u = e^x \), \( \frac{du}{dx} = e^x \). Thus, the derivative of \( \tan^{-1}(e^x) \) is \( \frac{e^x}{1 + (e^x)^2} = \frac{e^x}{1 + e^{2x}} \).
The derivative of \( f(x) \) is:
\( f'(x) = 2 \cdot \frac{e^x}{1 + e^{2x}} - 0 = \frac{2e^x}{1 + e^{2x}} \)
Since \( e^x>0 \) for all real \( x \), and \( 1 + e^{2x}>0 \), the derivative \( f'(x) = \frac{2e^x}{1 + e^{2x}} \) is always positive.
A positive derivative indicates that \( f(x) \) is strictly increasing over its entire domain \( (-\infty, \infty) \).
In summary, the function \( f(x) \) is odd and strictly increasing on \( (-\infty, \infty) \).