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Question Number 196309 by sonukgindia last updated on 22/Aug/23

Commented by Frix last updated on 22/Aug/23

$\sqrt{65} at x = - \frac{13}{11}$
	Answered by deleteduser1 last updated on 22/Aug/23
	$For x<−2∨x>0;f(x) is increasing ⇒min{f(x)} occurs at −2<x<0 ((2x+1)/( (√(2x^2 +2x+13))))+((2x+4)/( (√(2x^2 +8x+26))))=0 ⇒(2x+1)((√(2x^2 +8x+26)))+(2x+4)((√(2x^2 +2x+13)))=0 (2x+1)^2 (2x^2 +8x+26)=(2x+4)^2 (2x^2 +2x+13) ⇒(2x+1)^2 [(2x+4)^2 +36]=(2x+4)^2 [(2x+1)^2 +25] ⇒[6(2x+1)]^2 =[5(2x+4)]^2 ⇒12x+6=10x+20 or 12x+6=−10x−20⇒x=3 or x=((−13)/(11)) ⇒min{f(x)}=f(((−13)/(11)))=(√(65))$
	$F o r x < - 2 \lor x > 0; f (x) i s i n c r e a s i n g$ $\Rightarrow m i n {f (x)} o c c u r s a t - 2 < x < 0$ $\frac{2 x + 1}{\sqrt{2 x^{2} + 2 x + 13}} + \frac{2 x + 4}{\sqrt{2 x^{2} + 8 x + 26}} = 0$ $\Rightarrow (2 x + 1) (\sqrt{2 x^{2} + 8 x + 26}) + (2 x + 4) (\sqrt{2 x^{2} + 2 x + 13}) = 0$ ${(2 x + 1)}^{2} (2 x^{2} + 8 x + 26) = {(2 x + 4)}^{2} (2 x^{2} + 2 x + 13)$ $\Rightarrow {(2 x + 1)}^{2} [{(2 x + 4)}^{2} + 36] = {(2 x + 4)}^{2} [{(2 x + 1)}^{2} + 25]$ $\Rightarrow {[6 (2 x + 1)]}^{2} = {[5 (2 x + 4)]}^{2} \Rightarrow 12 x + 6 = 10 x + 20$ $o r 12 x + 6 = - 10 x - 20 \Rightarrow x = 3 o r x = \frac{- 13}{11}$ $\Rightarrow m i n {f (x)} = f (\frac{- 13}{11}) = \sqrt{65}$
	Answered by mr W last updated on 22/Aug/23

	$f (x) = \sqrt{2 {(x + \frac{1}{2})}^{2} + \frac{25}{2}} + \sqrt{2 {(x + 2)}^{2} + 18}$ $f (x) = \sqrt{2} [\sqrt{{(- x - \frac{1}{2})}^{2} + {(\frac{5}{2})}^{2}} + \sqrt{{(x + 2)}^{2} + 3^{2}}]$ $⩾ \sqrt{2} \times \sqrt{{(- x - \frac{1}{2} + x + 2)}^{2} + {(\frac{5}{2} + 3)}^{2}}$ $= \sqrt{\frac{3^{2} + 11^{2}}{2}} = \sqrt{65} = m i n i m u m$ $w h e n (- x - \frac{1}{2}) : (x + 2) = \frac{5}{2} : 3, i . e .$ $x = - \frac{13}{11}$
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