Question

$f(x) = \frac{1}{1 + x}$
হলে,
$f(f(f(x)))$
নির্ণয় কর।​

Answer 1

Answer:

$\bf \: given \\ \red{ \sf \: f(x) }= \frac{1}{ 1 + x} \\ \\ \bigstar \: \sf \: \orange{ f( \pink{f( \red{f(x)})}) }= \: to \: find \: \\ \\ \bf \: at \: \: first \: \\ \pink{f( \red{f(x)})} = \frac{1}{1 + \red{f(x)} } \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{1 + \frac{1}{1 + x} } \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{ \frac{1 + x + 1}{x + 1} } \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1 + x}{2 + x} \\ \\ \bf \: now \\ \bigstar \: \sf \: \orange{ f( \pink{f( \red{f(x)})}) } = \orange{f( \frac{1 + x}{2 + x} )} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{1 + \frac{1 + x}{ 2+ x} } \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{ \frac{2 + x + 1 + x}{2 + x} } \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2 +x}{3 + 2x} \\ \\ \therefore \: \: \sf \: \orange{ f( \pink{f( \red{f(x)})}) } = \frac{2 + x}{3 + 2x}$

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:f(x) = \dfrac{1}{1 + x}$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:f(f(f(x)))$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:f(x) = \dfrac{1}{1 + x}$

So,

$\rm :\longmapsto\:f(f(x)) = \dfrac{1}{1 + f(x)}$

$\rm :\longmapsto\:f(f(x)) = \dfrac{1}{1 + \dfrac{1}{1 + x} }$

$\rm :\longmapsto\:f(f(x)) = \dfrac{1}{ \: \: \dfrac{1 + x + 1}{1 + x} \: \: }$

$\rm :\longmapsto\:f(f(x)) = \dfrac{1}{ \: \: \dfrac{2 + x}{1 + x} \: \: }$

$\rm :\longmapsto\:f(f(x)) = \dfrac{1 + x}{2 + x }$

Now,

$\rm :\longmapsto\:f(f(f(x))) = \dfrac{1}{1 + f(f( x)) }$

$\rm :\longmapsto\:f(f(f(x))) = \dfrac{1}{1 + \dfrac{1 + x}{2 + x} }$

$\rm :\longmapsto\:f(f(f(x))) = \dfrac{1}{\dfrac{2 + x + 1 + x}{2 + x} }$

$\rm :\longmapsto\:f(f(f(x))) = \dfrac{1}{\dfrac{3 + 2x}{2 + x} }$

$\rm :\longmapsto\:f(f(f(x))) = \dfrac{2 + x}{3 + 2x}$

Additional Information :-

One - one :-

In order to show that f(x) is one - one, we have to choose two elements x and y belongs to domain such that f(x) = f(y), if on simplifying we get x = y, then f(x) is one - one otherwise its not one - one.

Onto :-

In order to show that f(x) is onto, we have to choose an element y belongs to co - domain such that f(x) = y. Then represent x as a function of g(y). If for every y, x exist, then f(x) is onto.