Question

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

Hint:

To demonstrate that a function is strictly increasing or decreasing, examine the sign of its derivative. If the derivative is always positive, the function is increasing.

Answer 1

Step 1: Differentiate the given function \( f(x) \).
The given function is: \[ f(x) = e^x - e^{-x} + x - \tan^{-1} x. \] Now, differentiate each term individually: \[ f'(x) = \frac{d}{dx} (e^x) - \frac{d}{dx} (e^{-x}) + \frac{d}{dx} (x) - \frac{d}{dx} (\tan^{-1} x). \] Thus, we obtain: \[ f'(x) = e^x + e^{-x} + 1 - \frac{1}{1 + x^2}. \] Step 2: Prove that \( f'(x)>0 \) for all \( x \).
Let's analyze the terms in \( f'(x) \): \[ f'(x) = e^x + e^{-x} + \frac{x^2}{1 + x^2}. \] - The terms \( e^x \) and \( e^{-x} \) are always positive for any real value of \( x \), as the exponential function never takes negative values. - Similarly, \( \frac{x^2}{1 + x^2} \) is always positive or zero, because \( x^2 \geq 0 \) for all real \( x \). Therefore, since all terms are positive: \[ f'(x)>0 \quad \text{for all } x. \] Step 3: Conclusion.
Since \( f'(x)>0 \) for every \( x \), the function \( f(x) \) is strictly increasing throughout its domain: \[ \boxed{\text{strictly increasing.}} \]