If f : R^+ → R^+ is a polynomial function which satisfy the equation f(f(x)) + f(x) = 12x ∀ x ∈ R^+ , then find f(x).


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Question Number 22133 by Tinkutara last updated on 11/Oct/17

$If f : R^{+} \to R^{+} is a polynomial function$ $which satisfy the equation f (f (x)) +$ $f (x) = 12 x \forall x \in R^{+}, then find f (x) .$
Commented by Rasheed.Sindhi last updated on 22/Oct/17
$f( f(x) )+f(x)=12x Case-1:f(x) is linear Let f(x)=ax+b f( f(x) )+f(x)=12x⇒ {a(ax+b)+b}+ax+b=12x a^2 x+ab+b+ax+b=12x (a^2 +a)x+ab+2b=12x Comparing coefficients: a^2 +a=12 ∧ ab+2b=0 (a−3)(a+4)=0 ∧ b(a+2)=0 (a=3 ∣ a=−4)∧ (b=0 ∣ a=−2) f(x)=3x, f(x)=−4x If f(x) is linear then f(x)=3x or f(x)=−4x$
$f (f (x)) + f (x) = 12 x$ $Case - 1 : f (x) is linear$ $Let f (x) = ax + b$ $f (f (x)) + f (x) = 12 x \Rightarrow$ ${a (ax + b) + b} + ax + b = 12 x$ $a^{2} x + ab + b + ax + b = 12 x$ $(a^{2} + a) x + ab + 2 b = 12 x$ $Comparing coefficients :$ $a^{2} + a = 12 \land ab + 2 b = 0$ $(a - 3) (a + 4) = 0 \land b (a + 2) = 0$ $(a = 3 ∣ a = - 4) \land (b = 0 ∣ a = - 2)$ $f (x) = 3 x, f (x) = - 4 x$ $If f (x) is linear then$ $f (x) = 3 x or f (x) = - 4 x$
	Answered by ajfour last updated on 14/Oct/17

	$l e t f (x) = y$ $A s f (y) + f (x) = 12 x$ $\Rightarrow \frac{d f}{d y} \times f^{'} (x) + f^{'} (x) = 12$ $o r (\frac{d f}{d y} + 1) f^{'} (x) = 12$ $\Rightarrow \frac{d f}{d y} = c o n s t a n t & f^{'} (x) = c o n s t a n t$ $l e t f^{'} (x) = a, t h e n \frac{d f}{d y} = \frac{12}{a} - 1 . . (i)$ $\Rightarrow f^{'} (x) = a x + b (. . s a y)$ $f (y) = a y + b$ $a n d \frac{d f}{d y} = a . . . (i i)$ $f r o m (i) a n d (i i) :$ $a = \frac{12}{a} - 1 \Rightarrow a^{2} + a - 12 = 0$ $\Rightarrow a = - 4, 3$ $n o w f (y) + f (x) = 12 x$ $\Rightarrow (a y + b) + (a x + b) = 12 x$ $\Rightarrow [a (a x + b) + b] + a x + b = 12 x$ $o r (a^{2} + a) x + b (a + 2) = 12 x$ $\Rightarrow a^{2} + a = 12 a n d b (a + 2) = 0$ $S o a = - 4, 3 w h i l e b = 0$ $H e n c e f (x) = - 4 x$ $o r f (x) = 3 x .$
	Commented by Tinkutara last updated on 14/Oct/17

	$Thank you very much Sir!$
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