Key Concepts and Formulas

• A function $f: A \to B$ is invertible if and only if it is both one-to-one (injective) and onto (surjective).
• A function $f$ is one-to-one (injective) if for any $x_1, x_2$ in the domain, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.
• A function $f: A \to B$ is onto (surjective) if for every $y$ in the codomain $B$, there exists at least one $x$ in the domain $A$ such that $f(x) = y$. This means the range of $f$ is equal to its codomain.
• If $f$ is invertible, its inverse function $f^{-1}: B \to A$ can be found by setting $y = f(x)$ and solving for $x$ in terms of $y$.

Step-by-Step Solution

Let the given function be $f: N \to Y$ defined as $f(x) = 4x + 3$. The domain is $N = \{1, 2, 3, \dots\}$. The codomain is $Y = \{ y \in N, y = 4x + 3 \text{ for some } x \in N \}$.

Step 1: Proving $f$ is One-to-One (Injective)

Reasoning: To show $f$ is one-to-one, we assume $f(x_1) = f(x_2)$ for two elements $x_1, x_2$ in the domain $N$ and prove that this implies $x_1 = x_2$.

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$. By the definition of $f(x)$, we have: $4x_1 + 3 = 4x_2 + 3$ Explanation: We substitute the function's rule into the assumption $f(x_1) = f(x_2)$.

Subtracting 3 from both sides of the equation: $4x_1 = 4x_2$ Explanation: This algebraic step isolates the terms involving $x_1$ and $x_2$.

Dividing both sides by 4: $x_1 = x_2$ Explanation: This final algebraic step shows that if the function values are equal, the domain values must also be equal.

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is one-to-one.

Step 2: Proving $f$ is Onto (Surjective)

Reasoning: To show $f$ is onto, we must demonstrate that for every element $y$ in the codomain $Y$, there exists an element $x$ in the domain $N$ such that $f(x) = y$. This means the range of $f$ must be equal to its codomain $Y$.

The codomain is given as $Y = \{ y \in N, y = 4x + 3 \text{ for some } x \in N \}$. Explanation: This definition itself states that $Y$ is the set of all possible output values of $f(x)$ when $x$ is a natural number.

Let $y$ be an arbitrary element in the codomain $Y$. By the definition of $Y$, if $y \in Y$, then there exists some $x_0 \in N$ such that $y = 4x_0 + 3$. Explanation: This means that for any element $y$ we pick from the codomain $Y$, we are guaranteed by its definition that it can be expressed in the form $4x_0 + 3$ for some natural number $x_0$. This $x_0$ is precisely the pre-image of $y$ in the domain $N$, as $f(x_0) = 4x_0 + 3 = y$.

Since every element $y$ in the codomain $Y$ has a pre-image $x_0$ in the domain $N$, the function $f$ is onto.

Conclusion for Invertibility: Since $f$ has been proven to be both one-to-one and onto, it is invertible.

Step 3: Finding the Inverse Function $g(y)$

Reasoning: To find the inverse function, we set $y = f(x)$ and algebraically solve for $x$ in terms of $y$. This expression for $x$ will be the inverse function, denoted by $g(y)$.

Let $y = f(x)$. $y = 4x + 3$ Explanation: We represent the output of the function $f(x)$ as $y$.

Our goal is to express $x$ in terms of $y$. Subtract 3 from both sides of the equation: $y - 3 = 4x$ Explanation: We isolate the term containing $x$ by moving the constant term to the other side.

Divide both sides by 4: $x = \frac{y - 3}{4}$ Explanation: This step isolates $x$, giving us its expression in terms of $y$.

Therefore, the inverse function $g(y)$ is given by: $g(y) = \frac{y - 3}{4}$

We can verify that for any $y \in Y$, $g(y) \in N$. Since $y \in Y$, $y = 4k+3$ for some $k \in N$. Then $g(y) = \frac{(4k+3)-3}{4} = \frac{4k}{4} = k$, which is in $N$.

Comparing our derived inverse function with the given options: (A) $g\left( y \right) = {{3y + 4} \over 4}$ (B) $g\left( y \right) = 4 + {{y + 3} \over 4}$ (C) $g\left( y \right) = {{y + 3} \over 4}$ (D) $g\left( y \right) = {{y - 3} \over 4}$

Our result matches option (D).

Common Mistakes & Tips

• Forgetting to prove invertibility: Always confirm that the function is both injective and surjective before proceeding to find the inverse.
• Algebraic errors when solving for x: Be meticulous with algebraic manipulations when isolating $x$ from the equation $y = f(x)$.
• Confusing domain and codomain of the inverse: Remember that the domain of $f^{-1}$ is the codomain of $f$, and the codomain of $f^{-1}$ is the domain of $f$.

Summary

To show that the function $f(x) = 4x + 3$ is invertible, we first proved it is one-to-one by showing $f(x_1) = f(x_2) \implies x_1 = x_2$. Then, we showed it is onto by confirming that for every $y$ in its defined codomain $Y$, there exists an $x$ in the domain $N$ such that $f(x) = y$. Finally, we found the inverse function by setting $y = 4x + 3$ and solving for $x$, yielding $g(y) = \frac{y - 3}{4}$.

The final answer is $\boxed{g\left( y \right) = {{y - 3} \over 4}}$, which corresponds to option (D).