Question

Which describes the behavior of the function f(x)= tan^-1(x)?

Answer 1

Answer:

Third Option

Step-by-step explanation:

We know that the function $tan(x)$ is defined as $y=\frac{sin(x)}{cos(x)}$. Since the denominator is $cos(x)$ then we know that  $cos(x)=0$  when $x=\frac{\pi}{2}$

We also know that the division by 0 is not defined. Therefore, the limit of $y=tan(x)$ when "x" tends to $\frac{\pi}{2}$ is infinite.

The function $tan^{-1}(x)$ is the inverse of $tan(x)$

By definition, if we have a function f(x), its domain will be equal to the range of its inverse function $f^{-1}(x)$. If $f(3)=8$, then $f^{-1}(8)=3$

This also happens for the function $tan^{-1}(x)$

If when $x \to \frac{\pi}{2}, tan(x) \to \infty$ then when $x \to \infty, tan^{-1}x \to \frac{\pi}{2}$

Then, the answer is:

$x \to \infty, f(x) \to \frac{\pi}{2}$