Question

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

Answer 1

The function f(x) = tan^-1(x) describes the arctangent, the inverse of the tangent function. It produces an angle whose tangent is the given x value, with a range of -pi/2 to pi/2 and properties of being odd, continuous, and monotonically increasing within the domain of 0 ≤ x ≤ 20.

Step-by-step explanation:

The behavior of the function f(x) = tan-1(x) describes the arctangent of x, which is the inverse function of the tangent. As an inverse trigonometric function, tan-1(x) will output an angle whose tangent is x. For values of x in the domain 0 ≤ x ≤ 20, arctangent will return an angle in radians that is in the range of -\(π/2\) to \(π/2\), which corresponds to the angles of -90 degrees to 90 degrees.

The graph of f(x) = tan-1(x) is a continuous curve that increases and approaches \(π/2\) as x approaches positive infinity and approaches -\(π/2\) as x approaches negative infinity. Since we're only concerned with 0 ≤ x ≤ 20, the function will be increasing and will lie between 0 and tan-1(20).

Answer 2

Third Option

Step-by-step explanation:

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

We also know that the division by 0 is not defined. Therefore, the limit of
`y=tan(x)` when "x" tends to
`(\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 (\pi)/(2), tan(x) \to \infty` then when
`x \to \infty, tan^(-1)x \to (\pi)/(2)`

Then, the answer is:

`x \to \infty, f(x) \to (\pi)/(2)`