The functions of each pair are inverse to each other. For each pair, check that both compositions give the identity function.
$F: \mathbb{R} \to \mathbb{R}$ and $F^{−1}:\mathbb{R} \to \mathbb{R}$ are defined by $F(x)=3x+2$ and $F^{−1}(y)=\dfrac{y−2}{3}$. for all y ∈ R
My attempt:
Inverse Function
For each particular but arbitrarily chosen $y \in \mathbb{R}$, according to the definition of $f^{-1}$, $f^{-1}(y) = \dfrac{y-2}{3}$ is a unique real number $x$ such that $f(x) = y$. \begin{align*} F(x) & = y\\ 3x + 2 & = y\\ x & = \frac{y-2}{3} \end{align*} Therefore, $f^{-1}(y) = \frac{y-2}{3}$.
Compositions of Functions
The functions $g \circ f$ and $f \circ g$ are defined as follows: $$(g \circ f)(x) = g(f(x)) = g(3x + 2) = 3x + 2$$ for all $x \in \mathbb{Z}$.
Let's consider some definitions.
Definition. Let $A$, $B$, and $C$ be sets. Let $f: A \to B$ and $g: B \to C$ be functions. The composition of functions $g$ and $f$ is the function $g \circ f: A \to C$ defined by $(g \circ f)(a) = g(f(a))$.
Definition. Let $f: A \to B$ be a function. Then $g: B \to A$ is said to be the inverse function of $f$ if \begin{align*} (g \circ f)(a) & = a & \text{for each $a \in A$}\\ (f \circ g)(b) & = b & \text{for each $b \in B$} \end{align*} If $g$ is the inverse of $f$, we write $g = f^{-1}$, where $f^{-1}$ is read $f$ inverse.
Let $f: \mathbb{R} \to \mathbb{R}$ be the function defined by $f(x) = 3x + 2$. Let $g: \mathbb{R} \to \mathbb{R}$ be the function defined by $g(y) = \frac{y - 2}{3}$. To show that $f$ and $g$ are inverses, we must show that $(g \circ f)(x) = x$ for each $x \in \mathbb{R}$ and that $(f \circ g)(y) = y$ for each $y \in \mathbb{R}$.
Let $x \in \mathbb{R}$. Then \begin{align*} (g \circ f)(x) & = g(f(x)) && \text{by definition}\\ & = g(3x + 2) && \text{substitute $3x + 2$ for $f(x)$}\\ & = \frac{3x + 2 - 2}{3} && \text{substitute $3x + 2$ for $x$ in the definition of $g$}\\ & = \frac{3x}{3} && \text{simplify}\\ & = x && \text{simplify} \end{align*} Hence, $(g \circ f)(x) = x$ for each $x \in \mathbb{R}$.
It remains to show that $(f \circ g)(y) = y$ for each $y \in \mathbb{R}$.
Let $y \in \mathbb{R}$. Then \begin{align*} (f \circ g)(y) & = f(g(y)) && \text{by definition}\\ & = f\left(\frac{y - 2}{3}\right) && \text{substitute $\frac{y - 2}{3}$ for $g(y)$}\\ & = 3\left(\frac{y - 2}{3}\right) + 2 && \text{substitute $\frac{y - 2}{3}$ for $x$ in the definition of $f$}\\ & = y - 2 + 2 && \text{simplify}\\ & = y && \text{simplify} \end{align*}