Consider the function f(x)=tanx with a domain of 0<x<π2, where x is in radians.
What is the range of f−1(x)?
- 0<x<∞
- −∞<x<∞
- 0<f−1(x)<∞
- −∞<f−1(x)<∞
- 0<x<π2
- 0<f−1(x)<π2
To find the range of the inverse function f^(-1)(x), we need to find the values that x can take on. In other words, we need to determine the possible values of f^(-1)(x) for all x in the domain of f(x).
The given function is f(x) = tan(x), with a domain of 0 < x < π/2. To find the inverse function, we need to solve the equation y = tan(x) for x.
To do this, we can take the inverse tangent (or arctan) of both sides of the equation:
arctan(y) = arctan(tan(x)).
Since the tangent function has a period of π, we can add or subtract any multiple of π to the angle x without changing the value of the tangent. So, we have:
arctan(y) = x + kπ,
where k is an integer.
Now, we need to determine the range of values that f^(-1)(x) can take on. Since the range of f(x) = tan(x) is all real numbers, the range of f^(-1)(x) will also be all real numbers.
Hence, the correct answer is: -∞ < f^(-1)(x) < ∞.
the range of f-1(x) is the domain of f(x)
I assume you mean π/2
consider the graph of tan-1x