max ∧ min of f(x) =(√x) +4(√((1−x)/2))


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Question Number 157116 by cortano last updated on 20/Oct/21

$m a x \land m i n o f f (x) = \sqrt{x} + 4 \sqrt{\frac{1 - x}{2}}$
Commented by cortano last updated on 20/Oct/21

$w i t h o u t d e r i v a t i v e$
Commented by mr W last updated on 20/Oct/21

$m a x = 3$ $m i n = 1$
	Answered by mr W last updated on 20/Oct/21
	$we see 0≤x≤1, f(x)≥0 so let x=sin^2 θ with 0≤θ≤(π/2) f(x)=sin θ+2(√2)cos θ=3 sin (θ+tan^(−1) 2(√2)) max=3 at θ=(π/2)−tan^(−1) 2(√2) or x=cos^2 (tan^(−1) 2(√2))=(1/9) min=min{f(0),f(1)}=f(1)=1$
	$w e s e e 0 ⩽ x ⩽ 1, f (x) ⩾ 0$ $s o l e t x = \sin^{2} θ w i t h 0 ⩽ θ ⩽ \frac{π}{2}$ $f (x) = \sin θ + 2 \sqrt{2} \cos θ = 3 \sin (θ + \tan^{- 1} 2 \sqrt{2})$ $m a x = 3 a t θ = \frac{π}{2} - \tan^{- 1} 2 \sqrt{2} o r x = \cos^{2} (\tan^{- 1} 2 \sqrt{2}) = \frac{1}{9}$ $m i n = m i n {f (0), f (1)} = f (1) = 1$
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