The function f : R → R is such that f ( x ) = x 2 + 1 x . Find f ( f ( 2 ) ) .
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you're answer is good about calculus but let me tell this is an alzebra so you try by that basic way to solve it as it wanted.
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this is not calculus. where you find this to be calculus?
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sorry ,thanks for informing mistake
Solve for f ( 2 ) ,
f ( 2 ) gives the value after putting the x = 2 is 2 / 5
f ( 2 ) = 2 / 5
then to find f ( f ( 2 ) )
put the value f ( 2 ) in given equation
f ( f ( 2 ) ) gives as
f ( f ( 2 ) ) = ( 2 / 5 ) / ( 2 / 5 ) ² + 1 = 1 0 / 2 9
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Put x = tan y So f ( tan y ) = 2 sin 2 y which yields f ( f ( tan y ) ) = sin 2 2 y + 4 2 sin 2 y . We know that tan y = 2 & sin 2 y = 5 4 which makes f ( f ( 2 ) ) = 1 1 6 4 0 = 2 9 1 0