Adjusted Composition

Algebra Level 2

$\large f(x) = \frac{\alpha x}{x+1}$

For $x\ne-1$ , consider a function $f$ as described above for some constant $\alpha$ . What is the value of $\alpha$ for which $f(f(x))$ is an identity function?

Clarification : $g(x)$ is an identity function if $g(x) = x$ .

Image Credit: Flickr Eliana Lúcio .

-1

-\sqrt 2

1

0

2 solutions

Kazem Sepehrinia
Jun 25, 2015

Let's find $\alpha$ such that $f(f(x))=\frac{\alpha \frac{\alpha x}{x+1}}{\frac{\alpha x}{x+1}+1}=x \\ \frac{\alpha^2 x}{(\alpha +1)x+1}=x \\ \alpha^2 x=(\alpha+1)x^2+x$ $\alpha=-1$ eliminates $(\alpha+1)x^2$ and equates $\alpha^2 x$ with $x$ .

To find the value of $\alpha$ the standard way we can just notice that ${ \alpha }^{ 2 }x=\left( \alpha +1 \right) { x }^{ 2 }+x\Rightarrow \left( \alpha +1 \right) \left( { x }^{ 2 }+x\left( 1-\alpha \right) \right) =0$ .

Personal Data - 5 years, 11 months ago

It would be great if someone could explain what are identical functions.Thankyou

Akhil Bansal - 5 years, 11 months ago

It was probably just misspelled seeing as $f\left( x \right)=x$ is an identity function .

Personal Data - 5 years, 11 months ago

I got this problem right but while I clicked on the problem, the answers got shuffled and inadvertently the wrong option was clicked.. My points got down a bit.. Is there any way to reverse this on brilliant?

Sanghamitra Anand - 5 years, 2 months ago

Akshay Bhatia
Jun 27, 2015

Its not an identical function, We rather call it an IDENTITY function!

Adjusted Composition

Image Credit: Flickr Eliana Lúcio .

2 solutions

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