Problem 32 - Calculus of a Single Variable

Question

Show that \(f\) is strictly monotonic on the given interval and therefore has an inverse function on that interval. \(f(x)=\cot x, \quad(0, \pi)\)

Step-by-Step Solution

Answer

The function \(f(x) = \cot x\) is strictly monotonic (specifically, strictly decreasing) on the interval \((0, \pi)\) and hence has an inverse function on that interval.

Step 1: Understand Definitions

Strictly monotonic refers to a function that continuously either increases or decreases, but not both, over a given interval. A function has an inverse if it is one-to-one, meaning that each x-value in the domain corresponds to exactly one y-value in the range and vice versa. In the context of this problem, proving a function is strictly monotonic on an interval will confirm that it has an inverse on that interval, as being strictly monotonic guarantees the function is one-to-one.

Step 2: Compute the Derivative of the Function

The derivative of the function \(f(x) = \cot x\) is \(f'(x) = -\csc^2 x\). The first derivative tells us how the function \(f(x)\) changes as \(x\) changes. In this sense, if \(f'(x)\) is positive, \(f(x)\) is increasing. If \(f'(x)\) is negative, \(f(x)\) is decreasing.

Step 3: Analyze the Sign of the First Derivative Over the Given Interval

Given that \(f'(x) = -\csc^2 x\), it's evident that \(f'(x)\) is always negative in the interval \((0, \pi)\) except for where it is undefined. Therefore, the function \(f(x)\) is strictly decreasing on the interval \((0, \pi)\).

Step 4: Conclude the Strict Monotonicity and the Existence of an Inverse Function

Given that \(f(x)=\cot x\) is strictly decreasing on the interval \((0, \pi)\), it means it's strictly monotonic (specifically, strictly decreasing) within the domain. Therefore, \(f(x) = \cot x\) has an inverse function on the interval \((0, \pi)\).