Let $$f: N \to Y$$ be a function defined as $$f(x) = 4x + 3$$, where $$Y = \{y \in N : y = 4x + 3$$ for some $$x \in N\}$$. Show that f is invertible and its inverse is

The function is $$f:N \rightarrow Y$$ defined by $$f(x)=4x+3$$, where $$N=\{1,2,3,\dots\}$$ and $$Y=\{\,y \in N \mid y=4x+3 \text{ for some } x\in N\}$$.

Step 1: Verify injectivity
Assume $$f(x_1)=f(x_2)$$. Then
$$4x_1+3 = 4x_2+3$$
$$\Rightarrow 4x_1 = 4x_2$$
$$\Rightarrow x_1 = x_2$$.
Hence $$f$$ is one-one (injective).

Step 2: Verify surjectivity
Take any $$y \in Y$$. By definition of $$Y$$, there exists some $$x \in N$$ such that
$$y = 4x+3 = f(x)$$.
Thus every element of $$Y$$ is an image of some element of $$N$$, so $$f$$ is onto (surjective).

Since $$f$$ is both injective and surjective, it is bijective, therefore invertible.

Step 3: Find the inverse rule
Start with $$y=f(x)=4x+3$$.
Solve for $$x$$:
$$4x = y-3 \;\;\Rightarrow\;\; x = \frac{y-3}{4}$$.
Define $$g:Y \rightarrow N$$ by
$$g(y)=\frac{y-3}{4}$$.

Step 4: Verify that $$g$$ is the inverse
1. $$g(f(x)) = g(4x+3)=\frac{(4x+3)-3}{4}=x$$ for every $$x\in N$$.
2. $$f(g(y))=f\!\left(\frac{y-3}{4}\right)=4\!\left(\frac{y-3}{4}\right)+3 = y$$ for every $$y\in Y$$.
Both compositions give the respective identity functions, so $$g=f^{-1}$$.

Therefore the inverse of $$f$$ is $$g(y)=\frac{y-3}{4}$$.

Option D which is: $$g(y)=\dfrac{y-3}{4}$$