1. Given function

We have $f:N \to Y, \qquad f(x)=4x+3$ where $Y=\{y\in N\mid y=4x+3 \text{ for some } x\in N\}.$

So the codomain $Y$ is exactly the set of all numbers of the form $4x+3$.

2. Show that $f$ is one-one

Suppose $f(x_1)=f(x_2).$ Then $4x_1+3=4x_2+3$ $\Rightarrow 4x_1=4x_2$ $\Rightarrow x_1=x_2.$

Hence, $f$ is injective (one-one).

3. Show that $f$ is onto

Take any $y\in Y$. By definition of $Y$, there exists some $x\in N$ such that $y=4x+3.$ But this means $y=f(x).$ So every element of $Y$ is the image of some element of $N$.

Hence, $f$ is surjective (onto).

4. Therefore, $f$ is invertible

Since $f$ is both one-one and onto, it is bijective, so it has an inverse.

5. Find the inverse

Let $y=4x+3.$ Solve for $x$: $y-3=4x$ $x=\frac{y-3}{4}.$

Therefore, $f^{-1}(y)=g(y)=\frac{y-3}{4}.$

6. Check with the options

• A: $\frac{3y+4}{4}$ — incorrect
• B: $4+\frac{y+3}{4}$ — incorrect
• C: $\frac{y+3}{4}$ — incorrect
• D: $\frac{y-3}{4}$ — correct

Also verify: $g(f(x))=\frac{(4x+3)-3}{4}=x$ and for $y\in Y$, $f(g(y))=4\left(\frac{y-3}{4}\right)+3=y.$

So the inverse is indeed $g(y)=\frac{y-3}{4}.$

7. Final answer

The correct option is D.