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(D) To show that $f$ is invertible,we need to find a function $g: Y 	o N$ such that $g \circ f = I_N$ and $f \circ g = I_Y$.Given $f(x) = 4x + 3$,let

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$g(y) = \frac{y - 3}{4}$

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(D) To show that $f$ is invertible,we need to find a function $g: Y 	o N$ such that $g \circ f = I_N$ and $f \circ g = I_Y$.Given $f(x) = 4x + 3$,let $y = f(x) = 4x + 3$.Solving for $x$ in terms of $y$,we get $y - 3 = 4x$,which implies $x = \frac{y - 3}{4}$.Define a function $g: Y 	o N$ by $g(y) = \frac{y - 3}{4}$.Now,check the composition $g \circ f(x) = g(f(x)) = g(4x + 3) = \frac{(4x + 3) - 3}{4} = \frac{4x}{4} = x = I_N(x)$.Next,check the composition $f \circ g(y) = f(g(y)) = f\left(\frac{y - 3}{4}ight) = 4\left(\frac{y - 3}{4}ight) + 3 = (y - 3) + 3 = y = I_Y(y)$.Since $g \circ f = I_N$ and $f \circ g = I_Y$,the function $f$ is invertible and its inverse is $g(y) = \frac{y - 3}{4}$.

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

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