If f x - a - x n 1- n - a -0 and n - N- then prove that f f x - x for all x-

We have, f(x)=(a x^n)^1/n, a gt;0 Now, f{f(x)}=f(a x^n)^1/n = [ a { (a x^n)^1/n}^n ]^1/n = [ a ( (a x^n) ) ]^1/n = [ a a+x^n ]^1/n =(x^n)^1/n =x ∴ f{f(x) ...

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Byju's Answer

Standard XII

Mathematics

Composite Function

If f x = a - ...

Question

If $f (x) = (a - x^{n})^{\frac{1}{n}}, a > 0$ and $n ϵ N$ , then prove that $f (f (x)) = x$ for all $x$ .

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Solution

We have,
$f (x) = (a - x^{n})^{\frac{1}{n}}, a > 0$
Now,
$f {f (x)} = f (a - x^{n})^{\frac{1}{n}}$
$= {[a - {(a - x^{n})^{\frac{1}{n}}}^{n}]}^{\frac{1}{n}}$
$= {[a - ((a - x^{n}))]}^{\frac{1}{n}}$
$= {[a - a + x^{n}]}^{\frac{1}{n}}$
$= (x^{n})^{\frac{1}{n}}$
$= x$
$∴ f {f (x)} = x$
Hence, proved.

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If f(x)=(a−xn)1n,a>0 and nϵN, then prove that f(f(x))=x for all x.

We have, f(x)=(a−xn)1n,a>0 Now, f{f(x)}=f(a−xn)1n =[a−{(a−xn)1n}n]1n =[a−((a−xn))]1n =[a−a+xn]1n =(xn)1n =x ∴f{f(x)}=x Hence, proved.

If $f (x) = (a - x^{n})^{\frac{1}{n}}, a > 0$ and $n ϵ N$ , then prove that $f (f (x)) = x$ for all $x$ .

We have,
$f (x) = (a - x^{n})^{\frac{1}{n}}, a > 0$
Now,
$f {f (x)} = f (a - x^{n})^{\frac{1}{n}}$
$= {[a - {(a - x^{n})^{\frac{1}{n}}}^{n}]}^{\frac{1}{n}}$
$= {[a - ((a - x^{n}))]}^{\frac{1}{n}}$
$= {[a - a + x^{n}]}^{\frac{1}{n}}$
$= (x^{n})^{\frac{1}{n}}$
$= x$
$∴ f {f (x)} = x$
Hence, proved.