The question asks whether the following inequality is true or false:

$-\frac{\pi}{2} < \tan^{-1}(-x) < \frac{\pi}{2}$

Solution

1. Understanding the Range of $\tan^{-1}$ (Arctan):

   • The function $\tan^{-1}(y)$ (or $\arctan(y)$) has a range of $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
   • This range means that for any real number $y$, the output of $\tan^{-1}(y)$ will always lie within $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.

2. Applying to $\tan^{-1}(-x)$:

   • Since $x$ is a real number, $-x$ is also real, and so $\tan^{-1}(-x)$ will also produce a value within the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

3. Conclusion:

   • The statement $-\frac{\pi}{2} < \tan^{-1}(-x) < \frac{\pi}{2}$ is true because it directly follows from the range of the $\tan^{-1}$ function.

Therefore, the answer is True.

Would you like further clarification or a more detailed explanation?

Related Questions:

1. What is the domain and range of the $\tan^{-1}$ function?
2. How does the $\tan^{-1}$ function behave for positive vs. negative inputs?
3. What other inverse trigonometric functions have restricted ranges?
4. How would the inequality change if it involved $\tan^{-1}(x)$ instead of $\tan^{-1}(-x)$?
5. What is the derivative of the $\tan^{-1}(x)$ function, and how is it used?

Tip:

Remember that inverse trigonometric functions have restricted ranges to ensure they are functions (one-to-one mappings). This is key when evaluating inequalities involving them.