max-min-of-f-x-x-4-1-x-2-

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

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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