Function(function(function))

Algebra Level 3

If f ( x ) = x x 2 + 1 f(x)=\frac{x}{\sqrt{x^{2}+1}} Find f ( f ( f ( x ) ) ) f(f(f(x))) .

None of the others x 2 x 2 + 1 \frac{x}{\sqrt{2x^{2}+1}} x 4 x 2 + 1 \frac{x}{\sqrt{4x^{2}+1}} x x 2 4 + 1 \frac{x}{\sqrt{\frac{x^{2}}{4}+1}} x x 2 + 1 \frac{x}{\sqrt{x^{2}+1}} x x 2 2 + 1 \frac{x}{\sqrt{\frac{x^{2}}{2}+1}}

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Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Omkar Kulkarni
May 20, 2015

I don't know if there's an easier way.

f ( x ) = x x 2 + 1 f(x)=\frac{x}{\sqrt{x^{2}+1}}

f ( f ( x ) ) = x x 2 + 1 ( x x 2 + 1 ) 2 + 1 = x x 2 + 1 2 x 2 + 1 x 2 + 1 = x 2 x 2 + 1 f(f(x))=\frac{\frac{x}{\sqrt{x^{2}+1}}}{\sqrt{\left(\frac{x}{\sqrt{x^{2}+1}}\right)^{2}+1}}=\frac{\frac{x}{\sqrt{x^{2}+1}}}{\frac{\sqrt{2x^{2}+1}}{\sqrt{x^{2}+1}}}=\frac{x}{\sqrt{2x^{2}+1}}

f ( f ( f ( x ) ) ) = x 2 x 2 + 1 ( x 2 x 2 + 1 ) 2 + 1 = x 2 x 2 + 1 3 x 2 + 1 2 x 2 + 1 = x 3 x 2 + 1 f(f(f(x)))=\frac{\frac{x}{\sqrt{2x^{2}+1}}}{\sqrt{\left(\frac{x}{\sqrt{2x^{2}+1}}\right)^{2}+1}}=\frac{\frac{x}{\sqrt{2x^{2}+1}}}{\frac{\sqrt{3x^{2}+1}}{\sqrt{2x^{2}+1}}}=\boxed{\frac{x}{\sqrt{3x^{2}+1}}}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

This is an interesting function, Omkar. If we define f n ( x ) f_{n}(x) as the function f f 'operated' n n times then we have that

f n ( x ) = x n x 2 + 1 . f_{n}(x) = \dfrac{x}{\sqrt{nx^{2} + 1}}.

Could make for another good question. :)

Brian Charlesworth - 6 years ago

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Sir your statement led me to do this . I hope you would like it.

Rohit Ner - 6 years ago

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Great question, Rohit. Definitely worth a reshare. :)

Brian Charlesworth - 6 years ago

Your comment led me to solve the question It's like retrosynthesis

Shashank Rustagi - 5 years, 12 months ago

PROOF PROOF PROOF PROOF!

Pi Han Goh - 6 years ago

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With all due respect , this cannot be proven because it is not even true!Could you please read my comment sir?

Arian Tashakkor - 6 years ago

There is an easier way actually.It becomes easier when you consider the fact that 2 x 2 + 1 x 2 + 1 \sqrt \frac {2x^2 + 1}{x^2+1} is not necessarily equal to 2 x 2 + 1 x 2 + 1 \frac {\sqrt {2x^2 + 1} }{\sqrt {x^2+1}} (Since the domains are not equal and we're talking about "Functions" here ... )

Arian Tashakkor - 6 years ago

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Over the reals I believe that those two expressions are always equal, as they are both necessarily positive. There would be "issues", (as there often are), if the domain were the complex numbers.

Brian Charlesworth - 6 years ago

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