Answer
$(F^{-1} \circ F)(x)=F^{-1}(F(x))=F^{-1}(3x+2)$
$\Rightarrow \frac{(3x+2)-2}{3}=\frac{3x}{3}=x=I_{\mathbb{R}}$
as was to be shown. Hence $F^{-1}\circ F=I_{\mathbb{R}}$ by definition of equality of functions.
$(F\circ F^{-1})(x)=F(F^{-1}(x))=F\left( \frac{y-2}{3} \right)$
$\Rightarrow 3\left( \frac{y-2}{3} \right)+2=y-2+2=y=I_{\mathbb{R}}$
as was to be shown. Hence $F\circ F^{-1}=I_{\mathbb{R}}$ by definition of equality of functions.
Work Step by Step
$(F^{-1} \circ F)(x)=F^{-1}(F(x))=F^{-1}(3x+2)$
$\Rightarrow \frac{(3x+2)-2}{3}=\frac{3x}{3}=x=I_{\mathbb{R}}$
as was to be shown. Hence $F^{-1}\circ F=I_{\mathbb{R}}$ by definition of equality of functions.
$(F\circ F^{-1})(x)=F(F^{-1}(x))=F\left( \frac{y-2}{3} \right)$
$\Rightarrow 3\left( \frac{y-2}{3} \right)+2=y-2+2=y=I_{\mathbb{R}}$
as was to be shown. Hence $F\circ F^{-1}=I_{\mathbb{R}}$ by definition of equality of functions.