If `f(x)=sec^(-1) x+ tan^(-1) x, " then " f(x)` is real for

Options

A. `x in [-1,1]`
B. `x in R`
C. `x in (- oo,1]uu[1+oo)`
D. none

Answer

To determine the values of \( x \) for which the function \( f(x) = \sec^{-1}(x) + \tan^{-1}(x) \) is real, we need to analyze the domains of the inverse functions involved.

### Step 1: Analyze the domain of \( \sec^{-1}(x) \)

The function \( \sec^{-1}(x) \) is defined for:
- \( x \leq -1 \) or \( x \geq 1 \)

This means that \( x \) must either be less than or equal to -1 or greater than or equal to 1 for \( \sec^{-1}(x) \) to be real.

### Step 2: Analyze the domain of \( \tan^{-1}(x) \)

The function \( \tan^{-1}(x) \) is defined for all real numbers \( x \). Therefore, there are no restrictions on \( x \) from this function.

### Step 3: Combine the domains

Since \( \tan^{-1}(x) \) is defined for all \( x \), the overall domain of \( f(x) \) will be determined solely by the domain of \( \sec^{-1}(x) \).

Thus, \( f(x) \) is real for:
- \( x \leq -1 \) or \( x \geq 1 \)

### Conclusion

The function \( f(x) = \sec^{-1}(x) + \tan^{-1}(x) \) is real for \( x \in (-\infty, -1] \cup [1, \infty) \).

---