((120)/(∣x−1∣ + ∣x−3∣ + ∣x−5∣)) ⇒ max = ?


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Question Number 206987 by hardmath last updated on 02/May/24

$\frac{120}{∣ x - 1 ∣ + ∣ x - 3 ∣ + ∣ x - 5 ∣} \Rightarrow \max = ?$
	Answered by Frix last updated on 02/May/24

	$\frac{120}{f (x)} is \max \Rightarrow f (x) is \min$ $4 ⩽∣ x - 1 ∣ + ∣ x - 3 ∣ + ∣ x - 5 ∣$ $\Rightarrow Answer is \frac{120}{4} = 30$
	Commented by hardmath last updated on 02/May/24

	$thank you very much dear professor$
	Answered by A5T last updated on 03/May/24

	$∣ x - 3 ∣=∣ 3 - x ∣; ∣ x - 5 ∣=∣ 5 - x ∣; ∣ a ∣ + ∣ b ∣⩾∣ a + b ∣$ $∣ x - 1 ∣ + ∣ 3 - x ∣⩾∣ x - 1 + 3 - x ∣= 2 . . . (i)$ $S i m i l a r l y, ∣ x - 1 ∣ + ∣ 5 - x ∣⩾ 4 . . . (i i)$ $∣ x - 3 ∣ + ∣ 5 - x ∣⩾ 2 . . . (i i i)$ $(i) + (i i) + (i i i) \Rightarrow∣ x - 1 ∣ + ∣ x - 3 ∣ + ∣ x - 5 ∣⩾ 4$ $(E q u a l i t y w h e n x = 3)$ $\Rightarrow \frac{120}{∣ x - 1 ∣ + ∣ x - 3 ∣ + ∣ x - 5 ∣} ⩽ \frac{120}{4} = 30 .$
	Commented by hardmath last updated on 03/May/24

	$thank you so much dear professor$
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