Problem 31 - 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. \(S=[-2,5), T=[-35,98), f(s)=s^{3}-27\)

Step-by-Step Solution

Answer

The function is invertible. The inverse is \( f^{-1}(t) = \sqrt[3]{t + 27} \).

Step 1: Define Invertibility

A function is invertible if it is both one-to-one (injective) and onto (surjective), meaning every element in the target set is mapped to by exactly one element in the domain.

Step 2: Check if f is One-to-One

For the function \( f(s) = s^3 - 27 \) to be one-to-one, each input \( s_1 eq s_2 \) needs to produce a different output \( f(s_1) eq f(s_2) \). For a cubic function, \( s^3 - 27 \), the function is strictly increasing in its domain \( S=[-2,5) \), hence it is one-to-one.

Step 3: Check Range of f

Determine the minimum and maximum values of \( f(s) \) on \( S=[-2,5) \).The minimum occurs at \( s = -2 \):\[ f(-2) = (-2)^3 - 27 = -8 - 27 = -35 \]The maximum occurs just before \( s = 5 \):\[ f(5) = 5^3 - 27 = 125 - 27 = 98 \]The range of \( f \) is thus \([-35, 98)\), equal to \( T \). Therefore, \( f \) is onto.

Step 4: Conclude Invertibility

Since \( f(s) = s^3 - 27 \) is both one-to-one and onto, \( f \) is invertible.

Step 5: Find Inverse Function Form

To find the inverse function, solve for \( s \) in terms of \( t \):\[ t = s^3 - 27 \]Add 27 to both sides:\[ t + 27 = s^3 \]Take the cube root:\[ s = \sqrt[3]{t + 27} \]So, the inverse function is: \( f^{-1}(t) = \sqrt[3]{t + 27} \)

Key Concepts

Injective Function

An injective function, also known as a one-to-one function, is a function where every element in the domain maps to a unique element in the codomain. This means that no two different inputs will produce the same output.

For instance, if we have two inputs, say \( s_1 \) and \( s_2 \) in the domain such that \( s_1 eq s_2 \), it is imperative that \( f(s_1) eq f(s_2) \) for the function to be injective. Injective functions are crucial because they ensure uniqueness in mapping, essentially guaranteeing that if two outputs are the same, then the inputs must have been identical.

In the original exercise, the given function \( f(s) = s^3 - 27 \) is verified to be injective over the domain \([-2,5)\) because it is strictly increasing. This characteristic of strictly increasing functions ensures that each different input results in a different output without overlaps.

Surjective Function

A surjective function, also called an onto function, is a situation where every element in the codomain has at least one matching element in the domain. In simpler terms, a function is surjective if its range is exactly the same as its codomain. This means that for every possible output, there is an input that maps to it.

In the exercise, the function \( f(s) = s^3 - 27 \) maps from the domain \([-2,5)\) to the codomain \([-35, 98)\). By calculating the function's values at the endpoints of the domain, we found that the range matches the codomain exactly. The minimum output \( -35 \) corresponds to the input \( s = -2 \), and the maximum output approaches \( 98 \) as \( s \) nears \( 5 \). Thus, the function is surjective.

Inverse Function

An inverse function essentially reverses the roles of inputs and outputs for a given function. For a function \( f \) to have an inverse, it must first be bijective. This means it needs to be both injective and surjective.

The process of finding an inverse involves solving the equation \( y = f(x) \) for \( x \) in terms of \( y \). In the original exercise, the function was \( f(s) = s^3 - 27 \). To find the inverse, the equation was solved for \( s \), resulting in \( f^{-1}(t) = \sqrt[3]{t + 27} \). This inverse function allows you to find the original input given an output, effectively "undoing" the function.

Cubic Function

A cubic function is a polynomial of degree three, and has the general form \( f(x) = ax^3 + bx^2 + cx + d \). The function we encountered in the exercise, \( f(s) = s^3 - 27 \), is a specific type of cubic function because its highest power of the variable \( s \) is 3, and it's simplified to have just two elements: a cubic term \( s^3 \) and a constant \(-27\).

Cubic functions can exhibit interesting properties when mapping, being flexible in showing different numbers of turning points or real roots depending on their coefficients. In this case, the function is strictly increasing over the given domain \([-2,5)\), which was instrumental in proving its injectivity and invertibility. Because it is monotonic throughout the domain, it avoids troublesome complexities such as having a maximum or minimum within the domain that could complicate the analysis of injectivity and surjectivity.