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

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

Commented bymnjuly1970 last updated on 06/Aug/21

thx master...

Commented bymnjuly1970 last updated on 06/Aug/21

min? (1/( (√2) ((√(2+(√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))))))