Question

If the function $f:R\to R$ be defined by $f\left(x\right)=2x-3andg:R\to R$ by $g\left(x\right)=x^3+5,$ then find fog and show that fog is invertible. Also find $(fog{)}^{-1},$ hence find $\left(fog{)}^{-1}\right(9).$

Answer 1

$fog:R\to R$ defined by $fog\left(x\right)=f\left(g\right(x\left)\right)=f(x^3+5)=2x^3+7=h\left(x\right)$ say.
Now $h\left(x\right)=2x^3+7.$
Let $x_1,x_2ϵR$ (Domain of h(x)) such that $h\left(x_1\right)=h\left(x_2\right)$
i.e., $2x_1^3+7=2x_2^3+7\Rightarrow x_1=x_2.$ Hence h(x) is one-one.
Let y = h(x) such that $yϵR$ (Codomain of h(x)).
That is $y=2x^3+7\Rightarrow x=(\frac{y-7}{2}{)}^{\frac{1}{3}}ϵR$ (Domain of h(x)).
So for every $yϵR$ there exists $xϵR$ such that $\left(\sqrt[3]{3\frac{y-7}{2}}\right)=y.$ Hence h(x) is onto.
As h(x) is one-one and onto so, it is invertible with $h^{-1}\left(x\right)=\left(fog{)}^{-1}\right(x)=\sqrt[3]{3\frac{x-7}{2}}.$
Also $\left(fog{)}^{-1}\right(9)=\sqrt[3]{3\frac{9-7}{2}}=1.$