Question

If $f\left(x\right)=\sqrt[n]{na-x^n},x>0,n\ge 2,n\in N$, then find the inverse of $f\left(x\right)$.

• A. $f^{-1}\left(x\right)=(a-x^n{)}^{1/n}$
• B. $f^{-1}\left(x\right)=a-x$
• C. $f^{-1}\left(x\right)=(a+x^n{)}^{1/n}$
• D. None of these

Answer 1

Answer: (A)

$f^{-1}\left(x\right)=(a-x^n{)}^{1/n}$

First method:
$f\left(x\right)=(a-x^n{)}^{1/n}$

$\left(fof\right)\left(x\right)=f\left[f\right(x\left)\right]$

$=f\left[\right(a-x^n{)}^{1/n}]$

$=(a-(f\left(x\right){)}^n{)}^{1/n}$

$=(a-a+x^n{)}^{1/n}=x$

$\Rightarrow \left(fof\right)\left(x\right)=x$

$\Rightarrow f^{-1}\left(x\right)=f\left(x\right)=\sqrt[n]{na-x^n}$
Alternative Method :
$f\left(x\right)=\sqrt[n]{na-x^n}$

Put $f\left(x\right)=y$

$y={(a-x^n)}^{1/n}$

Taking nth power ,

$y^n=a-x^n$

$x^n=a-y^n$

$\Rightarrow x={(a-y^n)}^{1/n}$

$\Rightarrow f^{-1}\left(y\right)={(a-y^n)}^{1/n}$

$\Rightarrow f^{-1}\left(x\right)={(a-x^n)}^{1/n}$