Max & min value of function f(x)=(√(6−x)) +(√(12+x)) .


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Question Number 153956 by liberty last updated on 12/Sep/21

$M a x & m i n v a l u e o f f u n c t i o n$ $f (x) = \sqrt{6 - x} + \sqrt{12 + x} .$
	Answered by EDWIN88 last updated on 12/Sep/21
	$domain of f(x) = [−12,6 ] by Chaucy −Schward inequality 2[(6−x)+(12+x) ]≥ ((√(6−x))+(√(12+x)) )^2 2×18 ≥ [ (√(6−x)) +(√(12+x)) ]^2 ⇒(√(18)) ≤ (√(6−x)) +(√(12+x)) ≤ (√(36)) ⇒3(√2) ≤ (√(6−x)) +(√(12+x)) ≤ 6 { ((min=3(√2))),((max=6)) :}$
	$d o m a i n o f f (x) = [- 12, 6]$ $b y C h a u c y - S c h w a r d i n e q u a l i t y$ $2 [(6 - x) + (12 + x)] ⩾ {(\sqrt{6 - x} + \sqrt{12 + x})}^{2}$ $2 \times 18 ⩾ {[\sqrt{6 - x} + \sqrt{12 + x}]}^{2}$ $\Rightarrow \sqrt{18} ⩽ \sqrt{6 - x} + \sqrt{12 + x} ⩽ \sqrt{36}$ $\Rightarrow 3 \sqrt{2} ⩽ \sqrt{6 - x} + \sqrt{12 + x} ⩽ 6$ ${\begin{cases} m i n = 3 \sqrt{2} \\ m a x = 6 \end{cases}$
	Commented by liberty last updated on 13/Sep/21

	$y e s c o r r e c t$
	Answered by ajfour last updated on 15/Sep/21

	$l e t x - 3 = t$ $f (t) = \sqrt{9 - t} + \sqrt{9 + t}$ $\Rightarrow t \in [- 9, 9]$ $f^{2} (t) = 18 + 2 \sqrt{81 - t^{2}}$ $c l e a r l y f_{m i n} = 3 \sqrt{2}, f_{m a x} = 6$
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