If $f(x) = \left( a - x^{n} \right)^{1/n}$, where $a 0$ and $n \in N$, then $fof(x)$ is equal to

Question

If $f(x) = \left( a - x^{n} \right)^{1/n}$, where $a > 0$ and $n \in N$, then $fof(x)$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given, $f(x) = \left( a - x^{n} ight)^{1/n} = g(x)$
$	herefore\ \ \ \ \ \ \ \ fof(x) = f\left( f(x) ight)$
$\ \ \ \ \ \ \  = \left\lbrack a - \left\{ \left( a - x^{n} ight)^{\frac{1}{n}} ight\}^{n} ightbrack^{1/n} = \left\lbrack a - \left( a - x^{n} ight) ightbrack^{1/n} = x$

Q. If $f(x) = \left( a - x^{n} \right)^{1/n}$, where $a > 0$ and $n \in N$, then $fof(x)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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