Question Number 12855 by kunalshukla95040 last updated on 04/May/17
![f(x)=[1−(x−3)^4 ]^(1/7) find f^(−1) (x).](https://www.tinkutara.com/question/Q12855.png)
$${f}\left({x}\right)=\left[\mathrm{1}−\left({x}−\mathrm{3}\right)^{\mathrm{4}} \right]^{\mathrm{1}/\mathrm{7}} \\ $$$${find}\:{f}^{−\mathrm{1}} \left({x}\right). \\ $$
Answered by linkelly0615 last updated on 04/May/17
![f(x)=[1−(x−3)^4 ]^(1/7) ⇒(f(x))^7 =1−(x−3)^4 ⇒1−(f(x))^7 =(x−3)^4 ⇒[1−(f(x))^7 ]^(1/4) =x−3 ⇒[1−(f(x))^7 ]^(1/4) +3=x ∵f^(−1) (f(x))=x ∴f^(−1) (x)=[1−x^7 ]^(1/4) +3](https://www.tinkutara.com/question/Q12856.png)
$${f}\left({x}\right)=\left[\mathrm{1}−\left({x}−\mathrm{3}\right)^{\mathrm{4}} \right]^{\mathrm{1}/\mathrm{7}} \\ $$$$\Rightarrow\left({f}\left({x}\right)\right)^{\mathrm{7}} =\mathrm{1}−\left({x}−\mathrm{3}\right)^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{1}−\left({f}\left({x}\right)\right)^{\mathrm{7}} =\left({x}−\mathrm{3}\right)^{\mathrm{4}} \\ $$$$\Rightarrow\left[\mathrm{1}−\left({f}\left({x}\right)\right)^{\mathrm{7}} \right]^{\mathrm{1}/\mathrm{4}} ={x}−\mathrm{3} \\ $$$$\Rightarrow\left[\mathrm{1}−\left({f}\left({x}\right)\right)^{\mathrm{7}} \right]^{\mathrm{1}/\mathrm{4}} +\mathrm{3}={x} \\ $$$$\because{f}^{−\mathrm{1}} \left({f}\left({x}\right)\right)={x} \\ $$$$\therefore{f}^{−\mathrm{1}} \left({x}\right)=\left[\mathrm{1}−{x}^{\mathrm{7}} \right]^{\mathrm{1}/\mathrm{4}} +\mathrm{3} \\ $$