min f(x)=(√(x^2 −2x+5)) +(√(4x^2 −4x+10))


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Question Number 175623 by cortano1 last updated on 04/Sep/22

$\min f (x) = \sqrt{x^{2} - 2 x + 5} + \sqrt{{4 x}^{2} - 4 x + 10}$
	Answered by greougoury555 last updated on 04/Sep/22
	$let { ((a=(1−x)i+2j)),((b=(2x−1)i+3j)) :} ∣a∣ +∣b ∣ ≥ ∣a +b∣ f(x) ≥ (√(x^2 +5^2 )) holds when ((1−x)/(2x−1)) = (2/3) 4x−2 = 3−3x ⇒x=(5/7) Min f(x)= (√(((25)/(49))+25)) = ((25(√2))/7)$
	$l e t {\begin{cases} a = (1 - x) i + 2 j \\ b = (2 x - 1) i + 3 j \end{cases}$ $∣ a ∣ + ∣ b ∣ ⩾ ∣ a + b ∣$ $f (x) ⩾ \sqrt{x^{2} + 5^{2}}$ $h o l d s w h e n \frac{1 - x}{2 x - 1} = \frac{2}{3}$ $4 x - 2 = 3 - 3 x \Rightarrow x = \frac{5}{7}$ $M i n f (x) = \sqrt{\frac{25}{49} + 25} = \frac{25 \sqrt{2}}{7}$
	Commented by infinityaction last updated on 04/Sep/22

	$w h a t i s i a n d j ? ?$
	Commented by mr W last updated on 04/Sep/22

	$i a n d j a r e u n i t v e c t o r s i n t h e p l a n e .$
	Commented by mr W last updated on 04/Sep/22

	$i t h i n k t h e s o l u t i o n i s w r o n g .$ $f r o m f (x) ⩾ \sqrt{x^{2} + 5^{2}}$ $y o u c a n n o t g e t t h e m i n i m u m o f f (x),$ $b e c a u s e \sqrt{x^{2} + 5^{2}} i s n o t a c o n s t a n t,$ $b u t a f u n c t i o n w h i c h i n t u r n h a s i t s$ $o n w m i n i m u m . y o u c a n n o t s a y$ $a t x = \frac{5}{7} \sqrt{x^{2} + 5^{2}} i s m i n i m u m .$
	Commented by infinityaction last updated on 04/Sep/22

	$t h a n k s s i r$
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