Question

Show that \[ f(x) = e^x - e^{-x} + x - \tan^{-1} x \] is strictly increasing in its domain.

Hint:

To prove monotonicity, show that the derivative is always positive (increasing) or negative (decreasing).

Answer 1

Step 1: Differentiate \( f(x) \).
The given function is: \[ f(x) = e^x - e^{-x} + x - \tan^{-1} x. \] Differentiate term by term: \[ f'(x) = \frac{d}{dx} (e^x) - \frac{d}{dx} (e^{-x}) + \frac{d}{dx} (x) - \frac{d}{dx} (\tan^{-1} x). \] This gives: \[ f'(x) = e^x + e^{-x} + 1 - \frac{1}{1 + x^2}. \] Step 2: Prove \( f'(x)>0 \).
Combine terms: \[ f'(x) = e^x + e^{-x} + \frac{x^2}{1 + x^2}. \] Since \( e^x>0 \), \( e^{-x}>0 \), and \( \frac{x^2}{1 + x^2}>0 \) for all \( x \), it follows that: \[ f'(x)>0 \quad \text{for all } x. \] Step 3: Conclusion.
Since \( f'(x)>0 \) for all \( x \), the function \( f(x) \) is strictly increasing in its domain: \[ \boxed{\text{strictly increasing.}} \]