f (x )= (1/( (√( 1 + sin (x ))) +(√( 1 + cos (x))))) find: Min( f (x)) =?


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Question Number 149667 by mnjuly1970 last updated on 06/Aug/21

$f (x) = \frac{1}{\sqrt{1 + s i n (x)} + \sqrt{1 + c o s (x)}}$ $f i n d :$ $Min (f (x)) = ?$
	Answered by iloveisrael last updated on 07/Aug/21
	$f(x)=(1/(∣cos (1/2)x+sin (1/2)x∣+(√2) ∣cos (1/2)x∣)) f(x)=(1/( (√2) ∣sin ((1/2)x+(π/4))∣+(√2) ∣cos (1/2)x∣)) when cos (1/2)x =1 or (1/2)x= 0 f_1 = (1/( (√2) sin ((π/4))+(√2))) = (1/( (√2) +1))=(√2)−1 when sin ((1/2)x+(π/4))=1 or (1/2)x=(π/4) f_2 = (1/( (√2) +(√2)((1/( (√2))))))=(1/( (√2)+1))=(√2)−1 f(x)= (√2) −1 when (1/2)x=(π/8) we get f_3 =(1/( (√2) {∣sin ((3π)/8)∣+∣cos (π/8)∣})) sin ((3π)/8)=(√((1−cos ((3π)/4))/2))=(√((1+((√2)/2))/2))=(√((2+(√2))/4))=((√(2+(√2)))/2) cos (π/8)=(√((1+cos (π/4))/2))=(√((2+(√2))/4))=((√(2+(√2)))/2) f(x)_(min) =f_3 =(1/( (√2) {((√(2+(√2)))/2)+((√(2+(√2)))/2)})) = (1/( (√2) ((√(2+(√2))))))$
	$f (x) = \frac{1}{∣ \cos \frac{1}{2} x + \sin \frac{1}{2} x ∣ + \sqrt{2} ∣ \cos \frac{1}{2} x ∣}$ $f (x) = \frac{1}{\sqrt{2} ∣ \sin (\frac{1}{2} x + \frac{π}{4}) ∣ + \sqrt{2} ∣ \cos \frac{1}{2} x ∣}$ $when \cos \frac{1}{2} x = 1$ $or \frac{1}{2} x = 0$ $f_{1} = \frac{1}{\sqrt{2} \sin (\frac{π}{4}) + \sqrt{2}} = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1$ $when \sin (\frac{1}{2} x + \frac{π}{4}) = 1$ $or \frac{1}{2} x = \frac{π}{4}$ $f_{2} = \frac{1}{\sqrt{2} + \sqrt{2} (\frac{1}{\sqrt{2}})} = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1$ $f (x) = \sqrt{2} - 1$ $when \frac{1}{2} x = \frac{π}{8} we get$ $f_{3} = \frac{1}{\sqrt{2} {∣ \sin \frac{3 π}{8} ∣ + ∣ \cos \frac{π}{8} ∣}}$ $\sin \frac{3 π}{8} = \sqrt{\frac{1 - \cos \frac{3 π}{4}}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$ $\cos \frac{π}{8} = \sqrt{\frac{1 + \cos \frac{π}{4}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$ $f {(x)}_{\min} = f_{3} = \frac{1}{\sqrt{2} {\frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 + \sqrt{2}}}{2}}}$ $= \frac{1}{\sqrt{2} (\sqrt{2 + \sqrt{2}})}$
	Commented by mnjuly1970 last updated on 06/Aug/21

	$t h x m a s t e r . . .$
	Commented by mnjuly1970 last updated on 06/Aug/21

	$m i n ? \frac{1}{\sqrt{2} (\sqrt{2 + \sqrt{2}})}$
	Answered by EDWIN88 last updated on 07/Aug/21
	$f(x)=(1/( (√(1+sin x))+(√(1+cos x)))) let g(x)=(√(1+sin x)) +(√(1+cos x)) then f(x)=(1/(g(x))) , f(x)_(min) it must be g(x)_(max) ⇒take g′(x)=((cos x)/(2(√(1+sin x))))−((sin x)/(2(√(1+cos x)))) =0 ⇒cos x(√(1+cos x)) = sin x(√(1+sin x)) ⇒cos^2 x+cos^3 x=sin^2 x+sin^3 x ⇒cos^2 x−sin^2 x=sin^3 x−cos^3 x ⇒(cos x−sin x)(cos x+sin x)=−(cos x−sin x)(1+sin xcos x) ⇒(cos x−sin x){cos x+sin x+1+sin xcos x)=0 when cos x=sin x ⇒x=(π/4) g(x)_(max) = (√(1+sin (π/4)))+(√(1+cos (π/4))) g(x)_(max) =(√((2+(√2))/2))+(√((2+(√2))/2)) g(x)_(max) =2(√((2+(√2))/2)) = (√2) ((√(2+(√2)))) therefore f(x)_(min) =(1/( (√2)((√(2+(√2))))))$
	$f (x) = \frac{1}{\sqrt{1 + \sin x} + \sqrt{1 + \cos x}}$ $let g (x) = \sqrt{1 + \sin x} + \sqrt{1 + \cos x}$ $t h e n f (x) = \frac{1}{g (x)}, f {(x)}_{m i n} i t$ $m u s t b e g {(x)}_{m a x}$ $\Rightarrow t a k e g^{'} (x) = \frac{\cos x}{2 \sqrt{1 + \sin x}} - \frac{\sin x}{2 \sqrt{1 + \cos x}} = 0$ $\Rightarrow \cos x \sqrt{1 + \cos x} = \sin x \sqrt{1 + \sin x}$ $\Rightarrow \cos^{2} x + \cos^{3} x = \sin^{2} x + \sin^{3} x$ $\Rightarrow \cos^{2} x - \sin^{2} x = \sin^{3} x - \cos^{3} x$ $\Rightarrow (\cos x - \sin x) (\cos x + \sin x) = - (\cos x - \sin x) (1 + \sin x \cos x)$ $\Rightarrow (\cos x - \sin x) {\cos x + \sin x + 1 + \sin x \cos x) = 0$ $w h e n \cos x = \sin x \Rightarrow x = \frac{π}{4}$ $g {(x)}_{m a x} = \sqrt{1 + \sin \frac{π}{4}} + \sqrt{1 + \cos \frac{π}{4}}$ $g {(x)}_{m a x} = \sqrt{\frac{2 + \sqrt{2}}{2}} + \sqrt{\frac{2 + \sqrt{2}}{2}}$ $g {(x)}_{m a x} = 2 \sqrt{\frac{2 + \sqrt{2}}{2}} = \sqrt{2} (\sqrt{2 + \sqrt{2}})$ $t h e r e f o r e f {(x)}_{m i n} = \frac{1}{\sqrt{2} (\sqrt{2 + \sqrt{2}})}$
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