Problem 30 - 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. \(S=(0, \infty), T=(1, \infty), f(s)=s^{4}+1\)

Step-by-Step Solution

Answer

The function is invertible; the inverse is \(f^{-1}(t) = (t - 1)^{1/4}\).

Step 1: Check if the function is one-to-one

A function is one-to-one if each element in the domain maps to a unique element in the codomain. For the function \(f(s) = s^4 + 1\), observe that if \(f(a) = f(b)\) implies \(a = b\), then the function is one-to-one. Let's compute:\[s^4 + 1 = t^4 + 1\] leads to \(s^4 = t^4\), which implies \(s = t\) because \(s, t > 0\). Therefore, \(f\) is one-to-one.

Step 2: Check if the function is onto

A function is onto if every element in the codomain \(T = (1, \, \infty)\) has a pre-image in the domain \(S = (0, \, \infty)\). For any \(t \in T\), \(t - 1 > 0\). Let \(s = (t - 1)^{1/4}\), then \(s^4 = t - 1\) and therefore \(f(s) = (t - 1) + 1 = t\). Hence, \(f\) is onto.

Step 3: Determine invertibility

Since the function \(f\) is both one-to-one and onto, it is invertible.

Step 4: Find the formula for the inverse

To find \(f^{-1}(t)\), solve \(t = s^4 + 1\) for \(s\). Rearrange to find \(s^4 = t - 1\), giving \(s = (t - 1)^{1/4}\). Thus, the inverse function is \(f^{-1}(t) = (t - 1)^{1/4}\).

Key Concepts

One-to-One Function

Understanding whether a function is one-to-one is crucial for determining its invertibility. A function is called one-to-one (or injective) if different inputs in its domain map to unique outputs in the codomain. In other words, if you pick any two distinct elements from the domain, their images need to be distinct as well. For the function in our example, specified as \(f(s) = s^4 + 1\), we need to check if \(f(a) = f(b)\) implies \(a = b\).

• Consider: \(f(a) = a^4 + 1\) and \(f(b) = b^4 + 1\).
• Set \(f(a) = f(b)\), giving \(a^4 + 1 = b^4 + 1\).
• This simplifies to \(a^4 = b^4\).
• Since \(a, b > 0\), it follows that \(a = b\).

Therefore, our function \(f(s) = s^4 + 1\) is one-to-one because only identical inputs produce identical outputs.

Onto Function

A function being onto (or surjective) implies that every element of the codomain is an image of at least one element from the domain. For a function \(f: S \rightarrow T\), where \(S = (0, \infty)\) and \(T = (1, \infty)\), we need to ensure that for every \(t\) in \(T\), there is some \(s\) in \(S\) such that \(f(s) = t\). Here is how we can test this:

• Assume an arbitrary \(t\) from \(T\).
• Set \(s = (t - 1)^{1/4}\).
• Compute \(s^4 = t - 1\), therefore \(f(s) = s^4 + 1 = t\).

By choosing \(s\) this way, every possible output in \(T\) is covered. Therefore, \(f(s) = s^4 + 1\) is onto.

Inverse Function

An inverse function reverses the roles of inputs and outputs such that applying the function and then its inverse returns the original input. In other words, if a function \(f\) is invertible, there exists an \(f^{-1}\) such that \(f(f^{-1}(t)) = t\) and \(f^{-1}(f(s)) = s\). Since our function \(f(s) = s^4 + 1\) has been found to be both one-to-one and onto, it is indeed invertible.
To find the inverse function, we must solve for \(s\) in terms of \(t\), starting from \(t = s^4 + 1\). Rearrange this to get \(s^4 = t - 1\).

• Taking the fourth root yields \(s = (t - 1)^{1/4}\).

Therefore, the inverse function is \(f^{-1}(t) = (t - 1)^{1/4}\). This function reverses the action of \(f\) by retrieving the original input from the output.

Function Domain

The domain of a function is the set of all possible input values it can accept. For our example, the domain \(S\) given is \((0, \infty)\). This means all positive real numbers are suitable inputs for the function \(f(s) = s^4 + 1\).
Understanding the domain is crucial for evaluating the behavior and possible outputs of a function. It helps to determine

• the range of values that the function can take,
• the validity of the operations involved in the function definition,
• and, in particular, it defines the region over which we test the injectivity and surjectivity of the function.

Since \(s^4 + 1\) operates on the positive reals, assertions about being one-to-one and onto hold within this defined domain, confirming the function's properties and ultimately its invertibility.