Csdrhrt

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 $y in Bbb{R}.$

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 $x in mathbb{Z}$.

If I understand well then you must check whether the functions $mathbb Rtomathbb R$ prescribed by $xmapsto3x+2$ and $xmapstofrac13(x-2)$ are inverses of each other.

That is indeed the case since their compositions are prescribed by: $xmapsto3x+2mapstofrac13((3x+2)-2)=x$ and: $xmapstofrac13(x-2)mapsto3(frac13(x-2))+2=x$ so both are identity functions.

This observation is enough to conclude that the functions are inverses of each other, and is what you are asked to do.