how do i prove this, help please. ∣((x^2 −2x−3)/(x^2 +2x+4))∣≤(5/3),∣x∣≤2


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Question Number 197275 by uchihayahia last updated on 12/Sep/23

$h o w d o i p r o v e t h i s, h e l p p l e a s e .$ $∣ \frac{x^{2} - 2 x - 3}{x^{2} + 2 x + 4} ∣⩽ \frac{5}{3}, ∣ x ∣⩽ 2$
	Answered by AST last updated on 12/Sep/23

	$O b s e r v e t h a t A = x^{2} + 2 x + 4 > 0 f o r a l l x a s i t h a s$ $n o p o s i t i v e r o o t s$ $\frac{d A}{d x} = 0 a t x = - 1 \Rightarrow m i n (A) = 3 \Rightarrow∣ x^{2} + 2 x + 4 ∣⩾ 3 . . (i)$ $∣ P ∣=∣ - P ∣\Rightarrow∣ x^{2} - 2 x - 3 ∣=∣ - x^{2} + 2 x + 3 ∣$ $\frac{d P}{d x} = 0 a t x = 1 \Rightarrow m a x (P) = 4$ $B u t a t x = - 2; P = - 5$ $\Rightarrow w h e n ∣ x ∣⩽ 2 (- 2 ⩽ x ⩽ 2) ∣ - x^{2} + 2 x + 3 ∣⩽ 5 . . . (i i)$ $\overset{?}{\Rightarrow} ∣ \frac{x^{2} - 2 x - 3}{x^{2} + 2 x + 4} ∣=∣ \frac{- x^{2} + 2 x + 3}{x^{2} + 2 x + 4} ∣⩽ \frac{5}{3}$ $(b u t e q u a l i t y c a n n o t h o l d, s i n c e (i) a n d (i i) o c c u r$ $a t d i f f e r e n t p o i n t s), s o q u e s t i o n i s n o t c o r r e c t$
	Commented by AST last updated on 12/Sep/23

	Commented by uchihayahia last updated on 13/Sep/23

	$i^{'} m s o r r y, b u t i c h e c k e d g e o g e b r a$ $a n d t h e t h e s t a t e m e n t i s t r u e$
	Commented by AST last updated on 13/Sep/23

	$S c r e e n s h o t ? A t w h a t p o i n t d o e s e q u a l i t y h o l d ?$
	Commented by uchihayahia last updated on 13/Sep/23


	Commented by uchihayahia last updated on 13/Sep/23

	$i {d o n}^{'} t k n o w w h y i {c a n}^{'} t s e n t a n i m a g e$ $a l w a y s g e t b l a n k$
	Commented by AST last updated on 13/Sep/23

	$W h a t p o i n t d o e s e q u a l i t y h o l d ?$
	Commented by uchihayahia last updated on 13/Sep/23

	$s a m e b u t t h e u p p e r b o u n d i s \frac{5}{4},$ ${i s n}^{'} t i t t r u e t o s a y \frac{5}{4} ⩽ \frac{5}{3}$ $∣ f (x) ∣⩽ \frac{5}{3} i s t r u e$
	Commented by AST last updated on 13/Sep/23

	$\frac{5}{4} < \frac{5}{3}; \frac{5}{4} \neq \frac{5}{3}$ $∣ f (x) ∣< \frac{5}{3} i s d i f f e r e n t f r o m ∣ f (x) ∣⩽ \frac{5}{3} b e c a u s e$ $e q u a l i t y n e v e r h o l d s i n t h e f o r m e r, w h i l e$ $e q u a l i t y h o l d s i n t h e l a t t e r .$
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