Question

Given $f:x\to 3x+6$ and $h:x\to \frac{x}{3}$, show that $\left(f^{-1}o{h}^{-1}\right)\left(x\right)=\left(hof{)}^{-1}\right(x)$

Answer 1

Since $f:x\to 3x+6$

so, $f^{-1}:x\to \frac{x-6}{3}$

and $h:x\to \frac{x}{3}$

so, $h^{-1}:x\to 3x$

$\Rightarrow f^{-1}\left(x\right)=\frac{x-6}{3}$

$f^{-1}\left(h^1\left(x\right)\right)=\left(f^{-1}0h^{-1}\right)\left(x\right)=\frac{h^{-1}\left(x\right)-6}{3}$

$\left(f^{-1}o{h}^{-1}\right)\left(x\right)=\frac{3x-6}{3}=x-2$

and $hof\left(x\right)=h\left(f\left(x\right)\right)=\frac{f\left(x\right)}{3}=\frac{3x+6}{3}=x+2$

since, $hof\left(x\right):x\to x+2$

so, ${\left(hof\right)}^{-1}\left(x\right):x\to x-2$

Hence, proved.