Problem 29 - 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. \(S=[0,1], T=[0,2], f(s)=s^{2}+s\)

Step-by-Step Solution

Answer

The function is invertible; the inverse is \( f^{-1}(t) = \frac{-1 + \sqrt{1 + 4t}}{2} \).

Step 1: Define the Concepts of One-to-One and Onto

A function is one-to-one (injective) if different inputs produce different outputs, i.e., \( f(a) = f(b) \Rightarrow a = b \) for all \( a, b \in S \). A function is onto (surjective) if every element in the codomain \( T \) has a pre-image in the domain \( S \). A function is invertible if it is both one-to-one and onto.

Step 2: Check if the Function is One-to-One

For \( f(s) = s^2 + s \), assume \( f(a) = f(b) \), then \( a^2 + a = b^2 + b \). Simplifying, \( a^2 - b^2 + a - b = (a-b)(a+b) + (a-b) = 0 \). Factor to get \( (a-b)(a+b+1) = 0 \), implying \( a = b \) if \( a + b + 1 eq 0 \). Since \( f \) is defined on \([0,1]\), it is one-to-one on the defined domain.

Step 3: Check if the Function is Onto

For \( y = f(s) = s^2 + s \), we must show that for every \( y \in [0,2] \), there exists an \( s \in [0,1] \) such that \( y = s^2 + s \). Set up the equation \( s^2 + s - y = 0 \). The discriminant of this quadratic equation is \( 1 + 4y \), which needs to be non-negative for real \( s \). For \( y \) in \( [0,2] \), \( 1 + 4y \) is non-negative, suggesting that \( f \) is onto.

Step 4: Confirm Invertibility and Find the Inverse

Since \( f \) is both one-to-one and onto, \( f \) is invertible. To find the inverse, solve \( y = s^2 + s \) for \( s \). Using the quadratic formula, \( s = \frac{-1 \pm \sqrt{1 + 4y}}{2} \). Since \( s \in [0,1] \), choose \( s = \frac{-1 + \sqrt{1 + 4y}}{2} \). Thus, \( f^{-1}(t) = \frac{-1 + \sqrt{1 + 4t}}{2} \).

Key Concepts

One-to-One (Injective)

Understanding if a function is one-to-one (or injective) is vital when determining if it's invertible. A function is considered injective if it never maps two different elements in its domain to the same element in its range. In simpler terms, every single input has a unique output.

Let's consider a function defined as \( f(s) = s^2 + s \) over the interval \( S = [0, 1] \). To check injectivity, assume \( f(a) = f(b) \). This leads to \( a^2 + a = b^2 + b \). Simplifying gives us \( (a-b)(a+b+1) = 0 \). For this product to be zero, either \( a = b \) or \( a + b + 1 = 0 \).

However, since the latter condition doesn't hold for \( a, b \) within \( [0,1] \), it shows that \( f(a) = f(b) \) only if \( a = b \). Therefore, the function is injective over this domain.

Onto (Surjective)

To check if a function is onto (surjective), we need to ensure that every possible output in the range can be linked to an input from the domain. This ensures that the entire target set (or codomain) is "covered" by the mapping.

For our function \( f(s) = s^2 + s \), defined from \( S = [0, 1] \) to \( T = [0, 2] \), we need to demonstrate that for each \( y \) in \( [0, 2] \), there exists a corresponding \( s \) in \( [0, 1] \) such that \( y = s^2 + s \).

We analyze the equation by setting \( s^2 + s - y = 0 \) and looking at its discriminant. This quadratic must have real solutions, which occurs when its discriminant, \( 1 + 4y \), is non-negative across the interval. For \( y \) within \( [0, 2] \), \( 1 + 4y \) remains non-negative, confirming the function's surjectivity.

Inverse Function

When a function is both one-to-one and onto, it is considered invertible. This means there exists another function that can effectively "undo" what the original function does. This inverse function can map each element of the range back to its original input from the domain.

For the function \( f(s) = s^2 + s \), confirmed as both injective and surjective, an inverse exists. To determine this inverse, we reverse the original function's mapping process: start with \( y = s^2 + s \) and solve for \( s \).

Using the quadratic formula, we find \( s = rac{-1 \pm \sqrt{1 + 4y}}{2} \). Given that \( s \) must lie within \( [0,1] \), take the positive root to maintain this condition. Therefore, the inverse function is denoted as \( f^{-1}(t) = \frac{-1 + \sqrt{1 + 4t}}{2} \). This formula allows any \( t \) within \( [0, 2] \) to be converted back to its corresponding \( s \) in \( [0, 1] \).