Function Of Function

Algebra Level pending

The function f : R R f : \mathbb {R \to R} is such that f ( x ) = x x 2 + 1 f(x) = \dfrac x {x^2 +1} . Find f ( f ( 2 ) ) f(f(2)) .

29 10 \frac{29}{10} 1 29 \frac{1}{29} 29 29 10 29 \frac{10}{29}

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2 solutions

Put x = tan y \displaystyle x=\tan y So f ( tan y ) = sin 2 y 2 \displaystyle f(\tan y) = \frac{\sin 2y}{2} which yields f ( f ( tan y ) ) = 2 sin 2 y sin 2 2 y + 4 \displaystyle f(f(\tan y)) = \frac{2\sin 2y}{\sin^2 2y+4} . We know that tan y = 2 \tan y=2 & sin 2 y = 4 5 \sin 2y =\frac{4}{5} which makes f ( f ( 2 ) ) = 40 116 = 10 29 \displaystyle f(f(2)) = \frac{40}{116}=\frac{10}{29}

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.

A Former Brilliant Member - 4 years, 9 months ago

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this is not calculus. where you find this to be calculus?

Aditya Narayan Sharma - 4 years, 9 months ago

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sorry ,thanks for informing mistake

A Former Brilliant Member - 4 years, 9 months ago

Solve for f ( 2 ) , f(2),

f ( 2 ) f(2) gives the value after putting the x = 2 x= 2 is 2 / 5 2/5

f ( 2 ) = 2 / 5 f(2)=2/5


then to find f ( f ( 2 ) ) f(f(2))

put the value f ( 2 ) f(2) in given equation

f ( f ( 2 ) ) f(f(2)) gives as

f ( f ( 2 ) ) = ( 2 / 5 ) / ( 2 / 5 ) ² + 1 = 10 / 29 f(f(2))=(2/5)/(2/5)²+1=10/29

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