Function Invertibility Analysis

To determine if a function ƒ : R → R defined by ƒ(x) = ax + b (where a, b are constants and a ≠ 0) is invertible, we must check if it is both one-to-one (injective) and onto (surjective). A function is invertible if and only if it is a bijection, meaning it is both injective and surjective.

One-to-One (Injectivity) Check

A function ƒ is considered one-to-one (or injective) if distinct elements in its domain always map to distinct elements in its codomain. In mathematical terms, for any \(x_1, x_2 \in R\), if \(\fnof;(x_1) = \fnof;(x_2)\), then it must follow that \(x_1 = x_2\).

Let's apply this condition to our given function ƒ(x) = ax + b:

• Assume that for two real numbers \(x_1\) and \(x_2\), \(\fnof;(x_1) = \fnof;(x_2)\).
• Substitute the definition of the function: \(ax_1 + b = ax_2 + b\).
• Subtract \(b\) from both sides of the equation: \(ax_1 = ax_2\).
• Since it is explicitly stated that \(a \ne 0\), we can safely divide both sides by \(a\): \(x_1 = x_2\).

Because \(\fnof;(x_1) = \fnof;(x_2)\) implies \(x_1 = x_2\), the function ƒ(x) = ax + b is indeed one-to-one.

Onto (Surjectivity) Check

A function ƒ : A → B is considered onto (or surjective) if every element in its codomain B has at least one corresponding element in its domain A. This means that for every \(y \in R\) (which is the codomain in this case), there must exist an \(x \in R\) (the domain) such that \(\fnof;(x) = y\).

Let's verify this for our function ƒ(x) = ax + b:

• Set the function equal to an arbitrary value \(y\) from the codomain: \(ax + b = y\).
• Our goal is to express \(x\) in terms of \(y\). First, isolate the term with \(x\): \(ax = y - b\).
• Now, solve for \(x\): \(x = \frac{y - b}{a}\).

Given that \(a\) and \(b\) are real constants and \(a \ne 0\), for any real number \(y\) you choose from the codomain R, the expression \(\frac{y - b}{a}\) will always result in a unique real number \(x\). This demonstrates that every \(y\) in the codomain has a pre-image \(x\) in the domain. Therefore, the function ƒ(x) = ax + b is onto.

Invertibility Conclusion

Since the function ƒ(x) = ax + b has been proven to be both one-to-one and onto, it satisfies the conditions for being a bijective function. As a direct consequence, any bijective function is always invertible.

Finding the Inverse Function ƒ^-1(x)

To determine the formula for the inverse function ƒ^-1(x), we typically follow these steps:

1. Step 1: Set \(y = \fnof;(x)\)
   • \(y = ax + b\)
2. Step 2: Solve the equation for \(x\) in terms of \(y\)
   • Subtract \(b\) from both sides: \(y - b = ax\)
   • Divide by \(a\) (since \(a \ne 0\)): \(x = \frac{y - b}{a}\)
3. Step 3: Swap \(x\) and \(y\) to express the inverse function in terms of \(x\)
   • Replace \(y\) with \(x\) and \(x\) with \(\fnof;^{-1}(x)\).
   • So, \(\fnof;^{-1}(x) = \frac{x - b}{a}\).

Thus, the inverse of the function ƒ(x) = ax + b is ƒ^-1(x) = \(\frac{x - b}{a}\). This result is characteristic of linear functions, where a non-zero slope (\(a \ne 0\)) ensures invertibility and a linear inverse function.