Funception

Algebra Level 3

If [ f ( x ) + x ] [ f ( x ) − x ] = 29 , f ( x ) ≥ 0 \left[ f\left( x \right) +x \right] \left[ f\left( x \right) -x \right] =29,f\left( x \right) \ge 0 ,find the value of f 139 ( 2015 ) { f }^{ 139 }(2015)

Note : f 2 ( x ) = ( f ∘ f ) ( x ) , f 3 ( x ) = ( f ∘ f ∘ f ) ( x ) f^{ 2 }\left( x \right) =\left( f\circ f \right) \left( x \right) ,f^{ 3 }\left( x \right) =\left( f\circ f\circ f \right) \left( x \right) and so on.


The answer is 2016.

Nud Mahachaiwong by Nud Mahachaiwong , 27, Thailand

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1 solution

Rishabh Jain
Dec 27, 2015

On simplyfying,we get- ( f ( x ) ) 2 − x 2 = 29 ⇒ f ( x ) = x 2 + 29 (f(x))^2 -x^2 =29 \Rightarrow f(x)=x^2+29 So f 2 ( x ) = √ x 2 + 29 ∗ 2 f^2 (x) =√{x^2+29*2} ; f 3 ( x ) = √ x 2 + 29 ∗ 3 f^3 (x) =√x^2+29*3 ..... f 139 ( x ) = √ x 2 + 29 ∗ 139 f^{139} (x) =√x^2+29*139 Therefore,

f 139 ( 2015 ) = √ 201 5 2 + 139 ∗ 29 f^{139}(2015)=√2015^2+139*29 =2016

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