Problem 37 - Calculus Single Variable

Question

A function \(f: S \rightarrow T\) is specified. Determine if \(f\) is invertible. If it is, state the formula for \(f^{-1}(t) .\) Otherwise, state whether \(f\) fails to be one-to-one, onto, or both. \(S=[1, \infty), T=(0,1], f(s)=1 /\left(s^{2}+1\right)\)

Step-by-Step Solution

Answer

The function is invertible, and \(f^{-1}(t) = \sqrt{\frac{1}{t} - 1}\).

Step 1: Check One-to-One Property

A function \(f\) is one-to-one if different inputs map to different outputs. To check this for the function \(f(s) = \frac{1}{s^2 + 1}\), assume \(f(a) = f(b)\). Then, \(\frac{1}{a^2 + 1} = \frac{1}{b^2 + 1}\). This implies \(a^2 + 1 = b^2 + 1\). Simplifying, we find that \(a = b\). Thus, \(f\) is one-to-one.

Step 2: Check Onto (Surjective) Property

For \(f: [1, \infty) \to (0,1]\) to be onto, every element \(t\) in \((0,1]\) must be the image of some \(s\) in \([1, \infty)\). The output \(f(s) = \frac{1}{s^2 + 1}\) approaches 1 as \(s \to 1\) and approaches 0 as \(s \to \infty\), covering the interval \((0,1]\). Thus, \(f\) is onto.

Step 3: Determine Invertibility

Since \(f\) is both one-to-one and onto, it is invertible. Now, we need to find the formula for \(f^{-1}(t)\). Given \(t = \frac{1}{s^2 + 1}\), solve for \(s\). Rearranging gives \(s^2 + 1 = \frac{1}{t}\) and \(s^2 = \frac{1}{t} - 1\). So, \(s = \sqrt{\frac{1}{t} - 1}\). Thus, \(f^{-1}(t) = \sqrt{\frac{1}{t} - 1}\).

Key Concepts

One-to-One Property

The one-to-one property of a function, also known as injectivity, means that each input corresponds to a distinct output. Imagine it as a unique 'fingerprint' for every input! In mathematical terms, if we consider two inputs, say \(a\) and \(b\), the function is one-to-one if \(f(a) = f(b)\) implies \(a = b\).
To determine if a given function \(f(s) = \frac{1}{s^2 + 1}\) is one-to-one, we assume that \(f(a) = f(b)\) and check if this results in \(a = b\). Starting with the equation \(\frac{1}{a^2 + 1} = \frac{1}{b^2 + 1}\), simplifying this shows \(a^2 + 1 = b^2 + 1\). This further reduces to \(a = b\).
Therefore, for this function, each unique input translates to a unique output, confirming that it is indeed one-to-one.

Onto Property

An onto function, also called surjective, suggests that every possible output from the codomain is hit by the function. Imagine trying to cover all targets with arrows and succeeding every time! For a function \(f: S \rightarrow T\) to be onto, each value \(t\) in the codomain \(T\) must have a corresponding input \(s\) in the domain \(S\).
In our example, the function \(f(s) = \frac{1}{s^2 + 1}\) maps from \([1, \infty)\) to \((0,1]\). As \(s\) approaches 1 from the right, \(f(s)\) nears the value 1. Conversely, as \(s\) becomes infinitely large, \(f(s)\) trends towards 0, effectively spanning the range of \((0,1]\).
This complete range coverage implies that for every \(t\) in \((0,1]\), there exists an \(s\) in \([1, \infty)\) such that \(f(s) = t\). Thus, the function is onto.

Inverse Function

An inverse function essentially 'reverses' the effect of the original function, mapping outputs back to their respective inputs. For a function to possess an inverse, it must be both one-to-one and onto.
Our task is to find \(f^{-1}(t)\) for the given \(f(s) = \frac{1}{s^2 + 1}\). Starting from the relationship \(t = \frac{1}{s^2 + 1}\), we rearrange it to solve for \(s\). First, rewrite the equation in terms of \(s^2 + 1\), yielding \(s^2 + 1 = \frac{1}{t}\). Subtracting 1 from both sides gives \(s^2 = \frac{1}{t} - 1\).
Consequently, taking the square root provides us with \(s = \sqrt{\frac{1}{t} - 1}\). Therefore, the inverse function is \(f^{-1}(t) = \sqrt{\frac{1}{t} - 1}\). This function allows us to backtrack from an output \(t\) to find its original input \(s\).

Function Mapping

Function mapping describes how elements from a domain are paired with elements in a codomain, forming 'input-output' pairs through the function rule.
Consider the function \(f(s) = \frac{1}{s^2 + 1}\), mapping elements from the domain \([1, \infty)\) to the codomain \((0,1]\). This relationship essentially tells us that any value \(s\) chosen from \([1, \infty)\) will become an output between 0 and 1 (not inclusive of 0) after being processed through the function.
Imagine the map as a series of arrows starting from each element in \([1, \infty)\) and pointing to a unique spot between 0 and 1, determined by the function formula \(f(s)\). This visual helps in understanding how an entire interval is transformed via function mapping, forming a bridge between two sets of numbers.