Find minimum and maximum value of function f(x)=(√(2sin x+3))−(√(sin x+1))


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 132531 by bemath last updated on 15/Feb/21

$Find minimum and maximum$ $value of function f (x) = \sqrt{2 \sin x + 3} - \sqrt{\sin x + 1}$
	Answered by liberty last updated on 15/Feb/21
	$((df(x))/dx)=((cos x)/( (√(2sin x+3))))−((cos x)/(2(√(sin x+1)))) =0 ((cos x)/( (√(2sin x+3)))) = ((cos x)/(2(√(sin x+1)))) cos^2 x(4sin x+4)=cos^2 x(2sin x+3) cos^2 (x) [ 2sin x+1 ]=0 { ((cos^2 x=0 → { ((x=(π/2))),((x=((3π)/2))) :})),((sin x=−(1/2)→x=((7π)/6))) :} (1)f((π/2))=(√(2sin (π/2)+3))−(√(sin (π/2)+1)) = (√5)−(√2) ≈ 0.8219 (2)f(((3π)/2))= (√(2sin ((3π)/2)+3))−(√(sin ((3π)/2)+1)) = (√1) −(√0) = 1 (maximum) (3) f(((7π)/6))=(√(2sin ((7π)/6)+3))−(√(sin ((7π)/6)+1)) = (√2) −(1/( (√2))) = (1/( (√2))) ≈ 0.7071 (minimum)$
	$\frac{df (x)}{dx} = \frac{\cos x}{\sqrt{2 \sin x + 3}} - \frac{\cos x}{2 \sqrt{\sin x + 1}} = 0$ $\frac{\cos x}{\sqrt{2 \sin x + 3}} = \frac{\cos x}{2 \sqrt{\sin x + 1}}$ $\cos^{2} x (4 \sin x + 4) = \cos^{2} x (2 \sin x + 3)$ $\cos^{2} (x) [2 \sin x + 1] = 0$ ${\begin{cases} \cos^{2} x = 0 \to {\begin{cases} x = \frac{π}{2} \\ x = \frac{3 π}{2} \end{cases} \\ \sin x = - \frac{1}{2} \to x = \frac{7 π}{6} \end{cases}$ $(1) f (\frac{π}{2}) = \sqrt{2 \sin \frac{π}{2} + 3} - \sqrt{\sin \frac{π}{2} + 1}$ $= \sqrt{5} - \sqrt{2} \approx 0.8219$ $(2) f (\frac{3 π}{2}) = \sqrt{2 \sin \frac{3 π}{2} + 3} - \sqrt{\sin \frac{3 π}{2} + 1}$ $= \sqrt{1} - \sqrt{0} = 1 (maximum)$ $(3) f (\frac{7 π}{6}) = \sqrt{2 \sin \frac{7 π}{6} + 3} - \sqrt{\sin \frac{7 π}{6} + 1}$ $= \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \approx 0.7071 (minimum)$
	Commented by liberty last updated on 15/Feb/21

	Answered by MJS_new last updated on 15/Feb/21

	$y = \sqrt{3 + 2 \sin x} - \sqrt{1 + \sin x}$ $let \sqrt{1 + \sin x} = u \land 0 ⩽ u ⩽ \sqrt{2}$ $y = \sqrt{2 u^{2} + 1} - u$ $\frac{d y}{d u} = \frac{2 u - \sqrt{2 u^{2} + 1}}{\sqrt{2 u^{2} + 1}} = 0 \Rightarrow u = \frac{\sqrt{2}}{2}$ $\Rightarrow$ $0 ⩽ u ⩽ \sqrt{2} \Rightarrow \frac{\sqrt{2}}{2} ⩽ y ⩽ 1$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum