f(x)=(((x+a)(x+b))/((x−a)(x−b))) Find minimum and maximum.


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Question Number 53732 by ajfour last updated on 25/Jan/19

$f (x) = \frac{(x + a) (x + b)}{(x - a) (x - b)}$ $F i n d m i n i m u m a n d m a x i m u m .$
Commented by mr W last updated on 25/Jan/19

$a t x = a a n d x = b : f (x) \to \pm \infty$
Commented by ajfour last updated on 25/Jan/19

$T h e r e i s a d e f i n i t e a n s w e r g i v e n,$ $m a x i m u m a n d m i n i m u m a r e$ $a s k e d . I t s f r o m a g o o d b o o k, S i r .$ $m u s t t h e n b e l o c a l m i n i m u m a n d$ $m a x i m u m .$
Commented by mr W last updated on 25/Jan/19

$s i n c e f (x) \to \pm \infty w h e n x \to a o r \to b,$ $t h e r e i s n o m i n . o r m a x . b u t$ $t h e r e c a n b e a l o c a l m i n i m u m a n d a$ $l o c a l m a x i m u m .$ $f (x) = \frac{x^{2} + (a + b) x + a b}{x^{2} - (a + b) x + a b}$ $f (x) = 1 + \frac{2 (a + b) x}{x^{2} - (a + b) x + a b}$ $f^{'} (x) = 0$ $\frac{1}{x^{2} - (a + b) x + a b} - \frac{x (2 x - (a + b))}{{(x^{2} - (a + b) x + a b)}^{2}} = 0$ $x^{2} - (a + b) x + a b = x (2 x - (a + b))$ $x^{2} = a b$ $\Rightarrow x = \pm \sqrt{a b}$ $i . e . i f a b ⩾ 0 t h e r e a r e l o c a l m i n . a n d$ $l o c a l m a x . a t x = \pm \sqrt{a b}$ $f {(x)}_{m i n / m a x} = \frac{(a \pm \sqrt{a b}) (b \pm \sqrt{a b})}{(a \mp \sqrt{a b}) (b \mp \sqrt{a b})}$ $i f a b < 0 t h e r e i s n o l o c a l m i n . o r m a x .$ $b u t w e {d o n}^{'} t n e e d t o u s e c a l c u l u s t o$ $g e t t h i s r e s u l t, j u s t l o o k a t$ $f (x) = 1 + \frac{2 (a + b) x}{x^{2} - (a + b) x + a b} = 1 + \frac{2 (a + b)}{x + \frac{a b}{x} - (a + b)}$ $i f a b ⩾ 0,$ $x + \frac{a b}{x} ⩾ 2 \sqrt{a b}$ $x + \frac{a b}{x} ⩽ - 2 \sqrt{a b}$ $f (x) = 1 + \frac{2 (a + b)}{x + \frac{a b}{x} - (a + b)} ⩽ 1 + \frac{2 (a + b)}{2 \sqrt{a b} - (a + b)} = - {(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}})}^{2} = f_{m a x}$ $f (x) = 1 + \frac{2 (a + b)}{x + \frac{a b}{x} - (a + b)} ⩾ 1 + \frac{2 (a + b)}{- 2 \sqrt{a b} - (a + b)} = - {(\frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}})}^{2} = f_{m i n}$
Commented by ajfour last updated on 25/Jan/19

$a n s w e r : - {(\frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}})}^{2} & - {(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}})}^{2} .$
Commented by mr W last updated on 25/Jan/19

$t h i s i s t h e s a m e a s m y r e s u l t (s e e b e l o w) .$ $a n s w e r i n b o o k i s n o t t h e b e s t, s i n c e$ $i t r e q u e s t s t h a t a ⩾ 0 a n d b ⩾ 0, b u t i n$ $f a c t t h i s i s n o t n e c e s s a r y . i t i s o n l y$ $n e c e s s a r y t h a t a b ⩾ 0, e . g . a = - 2, b = - 5 .$
Commented by mr W last updated on 25/Jan/19

$\frac{(a + \sqrt{a b}) (b + \sqrt{a b})}{(a - \sqrt{a b}) (b - \sqrt{a b})}$ $= \frac{\sqrt{a} (\sqrt{a} + \sqrt{b}) \sqrt{b} (\sqrt{b} + \sqrt{a})}{\sqrt{a} (\sqrt{a} - \sqrt{b}) \sqrt{b} (\sqrt{b} - \sqrt{a})}$ $= - \frac{(\sqrt{a} + \sqrt{b}) (\sqrt{a} + \sqrt{b})}{(\sqrt{a} - \sqrt{b}) (\sqrt{a} - \sqrt{b})}$ $= - {(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}})}^{2} = a s g i v e n i n b o o k$ $\frac{(a - \sqrt{a b}) (b - \sqrt{a b})}{(a + \sqrt{a b}) (b + \sqrt{a b})}$ $= \frac{\sqrt{a} (\sqrt{a} - \sqrt{b}) \sqrt{b} (\sqrt{b} - \sqrt{a})}{\sqrt{a} (\sqrt{a} + \sqrt{b}) \sqrt{b} (\sqrt{b} + \sqrt{a})}$ $= - \frac{(\sqrt{a} - \sqrt{b}) (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) (\sqrt{a} + \sqrt{b})}$ $= - {(\frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}})}^{2} = a s g i v e n i n b o o k$
Commented by ajfour last updated on 25/Jan/19

$y e s, s i r . t o o w e l l d i d y o u e x p l a i n .$ $T h a n k y o u t o o m u c h .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19
	$y=(((x+a)(x+b))/((x−a)(x−b))) lny=ln(x+a)+ln(x+b)−ln(x−a)−ln(x−b) (1/y)(dy/dx)=(1/(x+a))+(1/(x+b))−(1/(x−a))−(1/(x−b)) for max/min (dy/dx)=0 (1/(x+a))−(1/(x−b))+(1/(x+b))−(1/(x−a))=0 ((x−b−x−a)/((x+a)(x−b)))+((x−a−x−b)/((x+b)(x−a)))=0 ((−(a+b))/((x+a)(x−b)))+((−(a+b))/((x+b)(x−a)))=0 (1/((x+a)(x−b)))+(1/((x+b)(x−a)))=0 x^2 −ax+bx−ab+x^2 −bx+ax−ab=0 2x^2 −2ab=0 x=±(√(ab)) (1/y)(dy/dx)=(1/(x+a))+(1/(x+b))−(1/(x−a))−(1/(x−b)) (dy/dx)=(((x+a)(x+b))/((x−a)(x−b)))[(1/(x+a))+(1/(x+b))−(1/(x−a))−(1/(x−b))] ((wait...)/)wl now i am going to find change of sign using first derivativd method to find max/min ((dy/dx))=(((x+a)(x+b))/(x^2 −x(a+b)+ab))×[(1/(x+a))−(1/(x−b))+(1/(x+b))−(1/(x−a))] =((x^2 +x(a+b)+ab)/(x^2 −x(a+b)+ab))×[((x−b−x−a)/(x^2 −xb+ax−ab))+((x−a−x−b)/(x^2 −ax+bx−ab))] =(N/D)×[((−(a+b))/(x^2 −ab+x(a−b)))+((−(a+b))/(x^2 −ab−x(a−b)))] =(N/D)×−(a+b)×[((x^2 −ab−x(a−b)+x^2 −ab+x(a−b))/((x^2 −ab)^2 −x^2 (a−b)^2 ))] nowD<N so(N/D)>1 (N/D)×−(a+b)=−ve [((2(x^2 −ab))/((x^2 −ab)^2 −x^2 {(a+b)^2 −4ab))] when x>(√(ab)) x^2 =h+ab =(N/D)×−(a+b)×[((2(x^2 −ab))/((x^2 −ab)^2 −x^2 {(a+b)^2 −4ab))] =(−ve)×[((2h)/(h^2 −(h+ab)(a−b)^2 ))]=(−ve)×((+ve)/(−ve)) when x>(√(ab)) sign change from −ve to +ve so at x=(√(ab)) f(x) min y_ =(((x+a)(x+b))/((x−a)(x−b))) (at x=(√(ab)) ) =(((√a) ((√b) +(√a) )×(√b) ((√a) +(√b) ))/((√a) ((√b) −(√a) )×(√b) ((√a) −(√b) ))) =(−)((((√a) +(√b) )^2 )/(((√(b−))(√a) )^2 )) →min value so at x=(√(ab)) min value at x=−(√(ab)) max value y=(((−(√(ab)) +a)(−(√(ab)) +b))/((−(√(ab)) −a)(−(√(ab)) −b))) =(((√a) ((√a) −(√b) )×−(√b) ((√a) −(√b) ))/(−(√a) ((√a) +(√b) )×−(√b) ((√a) +(√b) ))) =(−1)×((((√a) −(√b) )^2 )/(((√a) +(√b) )))←max vzlud...$
	$y = \frac{(x + a) (x + b)}{(x - a) (x - b)}$ $l n y = l n (x + a) + l n (x + b) - l n (x - a) - l n (x - b)$ $\frac{1}{y} \frac{d y}{d x} = \frac{1}{x + a} + \frac{1}{x + b} - \frac{1}{x - a} - \frac{1}{x - b}$ $f o r m a x / m i n \frac{d y}{d x} = 0$ $\frac{1}{x + a} - \frac{1}{x - b} + \frac{1}{x + b} - \frac{1}{x - a} = 0$ $\frac{x - b - x - a}{(x + a) (x - b)} + \frac{x - a - x - b}{(x + b) (x - a)} = 0$ $\frac{- (a + b)}{(x + a) (x - b)} + \frac{- (a + b)}{(x + b) (x - a)} = 0$ $\frac{1}{(x + a) (x - b)} + \frac{1}{(x + b) (x - a)} = 0$ $x^{2} - a x + b x - a b + x^{2} - b x + a x - a b = 0$ $2 x^{2} - 2 a b = 0$ $x = \pm \sqrt{a b}$ $\frac{1}{y} \frac{d y}{d x} = \frac{1}{x + a} + \frac{1}{x + b} - \frac{1}{x - a} - \frac{1}{x - b}$ $\frac{d y}{d x} = \frac{(x + a) (x + b)}{(x - a) (x - b)} [\frac{1}{x + a} + \frac{1}{x + b} - \frac{1}{x - a} - \frac{1}{x - b}]$ $\frac{w a i t . . .}{} w l$ $n o w i a m g o i n g t o f i n d c h a n g e o f s i g n$ $u s i n g f i r s t d e r i v a t i v d m e t h o d t o f i n d m a x / m i n$ $(\frac{d y}{d x}) = \frac{(x + a) (x + b)}{x^{2} - x (a + b) + a b} \times [\frac{1}{x + a} - \frac{1}{x - b} + \frac{1}{x + b} - \frac{1}{x - a}]$ $= \frac{x^{2} + x (a + b) + a b}{x^{2} - x (a + b) + a b} \times [\frac{x - b - x - a}{x^{2} - x b + a x - a b} + \frac{x - a - x - b}{x^{2} - a x + b x - a b}]$ $= \frac{N}{D} \times [\frac{- (a + b)}{x^{2} - a b + x (a - b)} + \frac{- (a + b)}{x^{2} - a b - x (a - b)}]$ $= \frac{N}{D} \times - (a + b) \times [\frac{x^{2} - a b - x (a - b) + x^{2} - a b + x (a - b)}{{(x^{2} - a b)}^{2} - x^{2} {(a - b)}^{2}}]$ $n o w D < N s o \frac{N}{D} > 1 \frac{N}{D} \times - (a + b) = - v e$ $[\frac{2 (x^{2} - a b)}{{(x^{2} - a b)}^{2} - x^{2} {{(a + b)}^{2} - 4 a b}]$ $w h e n x > \sqrt{a b} x^{2} = h + a b$ $= \frac{N}{D} \times - (a + b) \times [\frac{2 (x^{2} - a b)}{{(x^{2} - a b)}^{2} - x^{2} {{(a + b)}^{2} - 4 a b}]$ $= (- v e) \times [\frac{2 h}{h^{2} - (h + a b) {(a - b)}^{2}}] = (- v e) \times \frac{+ v e}{- v e}$ $w h e n x > \sqrt{a b} s i g n c h a n g e f r o m - v e t o + v e$ $s o a t x = \sqrt{a b} f (x) m i n$ $y_{} = \frac{(x + a) (x + b)}{(x - a) (x - b)} (a t x = \sqrt{a b})$ $= \frac{\sqrt{a} (\sqrt{b} + \sqrt{a}) \times \sqrt{b} (\sqrt{a} + \sqrt{b})}{\sqrt{a} (\sqrt{b} - \sqrt{a}) \times \sqrt{b} (\sqrt{a} - \sqrt{b})}$ $= (-) \frac{{(\sqrt{a} + \sqrt{b})}^{2}}{{(\sqrt{b -} \sqrt{a})}^{2}} \to m i n v a l u e$ $s o a t x = \sqrt{a b} m i n v a l u e$ $a t x = - \sqrt{a b} m a x v a l u e$ $y = \frac{(- \sqrt{a b} + a) (- \sqrt{a b} + b)}{(- \sqrt{a b} - a) (- \sqrt{a b} - b)}$ $= \frac{\sqrt{a} (\sqrt{a} - \sqrt{b}) \times - \sqrt{b} (\sqrt{a} - \sqrt{b})}{- \sqrt{a} (\sqrt{a} + \sqrt{b}) \times - \sqrt{b} (\sqrt{a} + \sqrt{b})}$ $= (- 1) \times \frac{{(\sqrt{a} - \sqrt{b})}^{2}}{(\sqrt{a} + \sqrt{b})} \leftarrow m a x v z l u d . . .$
	Commented by ajfour last updated on 25/Jan/19

	$w e l c o m e b a c k S i r . .$
	Commented by ajfour last updated on 25/Jan/19

	$t h a n k s s i r, d o i p o s t t h e a n s w e r ?$
	Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

	$i w e n t t o s e e t h e n e w s . . .$
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